Order the numbers from least to greatest based on their absolute values.

|23|, |−37|, |−6|, |18|, |−24|, |2|

Answer:

C

Step-by-step explanation:

2/3x - 5>3

Add 5 to each side

2/3x - 5+5>3+5

2/3x > 8

Multiply each side by 3/2

3/2 *2/3x > 8*3/2

x > 12

There is an open circle at 12 and the lines goes to the right

Answer:

6e+52

Step-by-step explanation:

cAlCuLaToR

Answer:

6e+52

Step-by-step explanation:

multiply

Answer:

D. 150°

Step-by-step explanation:

x= 150°

Choice D

Answer:

0.6042 or 29/48

Step-by-step explanation:

-5/16 = -0.3125

7/24 = 0.2917

0.2917 - -0.3125 = 0.6042

0.6042 ≅ 29/48

Answer:

29/48

Step-by-step explanation:

-5/16 + x= 7/24

x= 7/24-(-5/16)

x=7/24+5/16

x= 2*7/2*24+ 3*5/3*16

x=29/48

Answer:

B:

Step-by-step explanation:

According to theorem, "the angle in a semi-circle is a right angle" So,

<O = 90°

<M = 54

<K = 180-90-54

<OKM = 36°

Answer:

40 litres

Step-by-step explanation:

V = l x w x h

50 x 40 x 20 = 40000

40000 cm^3

1cm^3 = 1ml

40000 cm^3/ 1cm^3 = 40000ml

40000 x 10^-3 = 40 litres

Answer:

The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

For the third class 78 - 83 ; the upper limit = 83

The class width for this histogram 5

The number of students that took the exam simply refers to the frequency is  84

The percentage of students that scored higher than 95.5 is 53%

Step-by-step explanation:

The objective of this question is to use the following histogram that shows the exam scores for a Pre-algebra class to answer the question given:

NOW;

The table given in the question can be illustrated as follows:

S/N           Class                Score          Frequency

1                 68  - 73            70.5           0

2                73 - 78             75.5           4

3                78 - 83             80.5           8

4                83 - 88             85.5           12

5                88 - 93            90.5           16

6                93 - 98            95.5           20

7                98 - 103           100.5          24                            

TOTAL:                                                 84

a) The number of the class containing the  largest score can be found in frequency  24 and the class is  98 - 103

b) For the third class 78 - 83 ; the upper limit = 83  ( since the upper limit is derived by addition of  5 to the last number showing  in the highest value specified by the number in the class interval which is 78 ( i.e 78 + 5 = 83))

c)   The class width for this histogram 5 ; since it  is the difference  between the upper and lower boundaries  limit of the given class.

So , from above the difference in any of the class will definitely result into 5

d) The number of students that took the exam simply refers to the frequency ; which is (0+4+8+12+16+20+24) = 84

e) Lastly; the percentage of students that scored higher than 95.5 is ;

⇒[tex]\dfrac{20+24}{84} *100[/tex]

= 0.5238095 × 100

= 52.83

To the nearest percentage ;the percentage of students that scored higher than 95.5 is 53%

Answer:

1. 98-103 (6th class)

2. 88

3. 5

4. 84

5. 52%

Step-by-step explanation:

Find attached the frequency table.

The class of exam scores falls between (1, 2, 3, 4, 5, or 6).

The exam score ranged from 68-103

1) The largest number of exam scores = 24

The largest number of exam scores is in the 6th class = 98 -103

Step 2 of 5:

The upper class limit is the higher number in an interval. Third class interval is 83-88

The upper class limit of the third class 88.

Step 3 of 5:

Class width = upper class limit - lower class limit

We can use any of the class interval to find this as the answer will be the same. Using the interval between 73-78

Class width = 78 - 73

Class width for the histogram = 5

Step 4 of 5:

The total of students that took the test = sum of all the frequency

= 0+4+8+12+16+20+24 = 84

The total of students that took the test = 84

Step 5 of 5:Find the percentage of students that scored higher than 95.5

Number of student that scored higher than 95.5 = 20 + 24 = 44

Percentage of students that scored higher than 95.5 = [(Number of student that scored higher than 95.5)/(total number of students that took the test)] × 100

= (44/84) × 100 = 0.5238 × 100 = 52.38%

Percentage of students that scored higher than 95.5 = 52% (nearest percent)

Answer:

x<_ 21

Step-by-step explanation:

5(x+27)>_ =6(x+26)

5x +135 >_ 6x +156

5x >_6x +21

-x>_21

x<_21

Answer:

63 metres

Step-by-step explanation:

A rectangle has 4 sides

2 of these sides are the lengths

The other 2 sides are the width

If the length of one side is 97 metres, the other side length must also be 97 metres

The two lengths then add together (97 + 97) to become 194 metres

Now we can use this information to calculate the width

320 (the total perimeter) subtract 194 (The total length) = 126 metres

This means that 126 metres is the total width

Because there are two sides which add up to the total width we divide 126 by 2

This allows us to get the measurement of the width

126 divided by 2 = 63 metres

Answer

The figure shows three lines that intersect at point N.

3 lines intersect. Lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees.

Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL?

42°

48°

132°

138°

Step-by-step explanation:

42

The required measure of angle MNL is 42°. Option A is correct.

3 lines intersect lines G K and J M intersect at point N to form a right angle. Line H L intersects the other lines at point N. Angle H N G is 48 degrees. Angle GNH is congruent to angle KNL. Angle MNL is complementary to angle KNL. What is the measure of angle MNL is to determine

The angle can be defined as the one line inclined over another line.
unit of measure of an angle is degree and radians.

Angle GNH is congruent to angle KNL.
∠KNL = 48°
Angle MNL is complementary to angle KNL
Since angle MNK = 90°
∠MNL + ∠KNL = 90°

∠MNL = 90-48
∠MNL = 42°

Thus, the required measure of angle MNL is 42°. Option A is correct.

Learn more about Angles here:
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Answer:

90° and 180°

Step-by-step explanation:

An obtuse angle is any angle larger than 90° and smaller than 180°

Answer:

The answer is 8

Step-by-step explanation:

Plug -2 in for x. The double-negative inside the parenthesis makes it positive, then do the exponent.

Answer:

-(-2)^3 = 2^3 = 8

Answer is C

Step-by-step explanation:

So we plug in the numbers. We have -2 as x. (-(-2)^3 would be our thing. Thats  because our x is the negative so the negative of -2 is 2.

2^3 = 8

therefore its 8

Tell me if it is right

Answer:

213, 123

Step-by-step explanation:

Answer:

The Pythagorean Theorum

Step-by-step explanation:

you can literally search it and see that it is right!

Answer:

The test statistic to test the null hypothesis equals 1.059

Step-by-step explanation:

From the given information; we have:

Treatment               Observations

A                                20        30         25       33

B                                22         26        20       28

C                               40         30         28       22

The objective is to find the  test statistic to test the null hypothesis; in order to do that;we must first run through a series of some activities.

Let first compute the sum of the square;

Total sum of squares (TSS) = Treatment sum of squares [tex](T_r SS)[/tex]  +    Error sum of squares   (ESS)

where:

(TSS) = [tex]\sum \limits ^v_{i=1} \sum \limits ^{n_i}_{j-1}(yij- \overline {y}oo)^2[/tex]  with (n-1)   df

[tex](T_r SS)[/tex]   [tex]= \sum \limits ^v_{i=1} n_i( \overline yio- \overline {y}oo)^2[/tex]   with (v-1)  df

[tex](ESS) = \sum \limits ^v_{i=1} \sum \limits ^{n_i}_{j-1}(yij- \overline {y}io)^2[/tex]   with (n-v) df

where;

v= 3

[tex]n_i=[/tex]4

i = 1,2,3

n =12

[tex]y_{ij}[/tex] is the [tex]j^{th[/tex] observation for the [tex]i^{th[/tex] treatment

[tex]\overline{y}io[/tex] is the mean of the  [tex]i^{th[/tex] treatment  i = 1,2,3 ;  j = 1,2,3,4

[tex]\overline y oo[/tex]   is the overall mean

From the given data

[tex]\overline y oo = \dfrac{1}{12} \sum \limits ^3_{i=1} \sum \limits ^{4}_{j=1}(yij)^2= 27[/tex]

[tex]TSS = \dfrac{1}{12} \sum \limits ^3_{i=1} \sum \limits ^{4}_{j=1}(yij- 27)^2 = 378[/tex]

[tex]T_r SS= \sum \limits^3_{i=1}4 (\overline y io - \overline yoo)^2[/tex]

[tex]=4(27-27)^2+4(24-27)^2+4(30-27)^2 = 72[/tex]

Total sum of squares (TSS) = Treatment sum of squares [tex](T_r SS)[/tex]  +    Error sum of squares   (ESS)

(TSS) =  378 - 72

(TSS) =  306

Now; to the mean square between treatments (MSTR); we use the formula:

MSTR = TrSS/df(TrSS)

MSTR = 72/(3 - 1)

MSTR = 72/2

MSTR = 36

The mean square within treatments  (MSE) is:

MSE = ESS/df(ESS)

MSE = 306/(12-3)

MSE = 306/(9)

MSE = 34

The test statistic to test the null hypothesis is :

[tex]T = \dfrac{ \dfrac{TrSS}{\sigma^2}/(v-1) }{ \dfrac{ESS}{\sigma^2}/(n-v) } = \dfrac{MSTR}{MSE} \ \ \ \approx \ \ T(\overline {v-1}, \overline {n-v})[/tex]

[tex]T = \dfrac{36}{34}[/tex]

T = 1.059

Answer:

A

Step-by-step explanation:

They are congruent because of the SSS theorem.  The chords are congruent because the angles are congruent (angles are congruent because they are vertical angles).  Congruent central angles have congruent chords.  The other two sides are congruent because they are all radii of the circle and radius are always congruent.

Answer:

A

Step-by-step explanation:

The triangles are isosceles and congruent  too. ( both have 2  sides congruent  as  radius and the angles between them are congruent- SAS)

Answer:

(7, 5) is the final reflection of the point.

Step-by-step explanation:

We are given point A(-7, 5) which is first reflected over the line [tex]x= -5[/tex].

The minimum distance of the point A(-7, 5) from the line [tex]x= -5[/tex] is 2 units across the horizontal path (No change in y coordinate).

Point A lies 2 units on the left side of the line [tex]x= -5[/tex].

So, its reflection will be 2 units on the right side of [tex]x= -5[/tex].

Let its reflection be A' which has coordinates as (-5+2,5) i.e. (-3, 5).

Now A'(-3, 5) is reflected on the line [tex]x=2[/tex].

The minimum distance of the point A'(-3, 5)  from the line [tex]x=2[/tex] is 5 units across the horizontal path (No change in y coordinate).

Point A' lies 5 units on the left side of the line [tex]x=2[/tex].

So, its reflection will be 5 units on the right side of [tex]x=2[/tex].

Let its reflection be A'' which has coordinates as (2+5, 5) i.e (7, 5) is the final reflection of the point..

Please find attached image.

(7, 5) is the final reflection of the point.

Answer:C

Step-by-step explanation:

the x value  5  corresponds to two difference y-values.

Answer:

f\left(x\right)=x^3-x

Step-by-step explanation:

Answer: There is a 0.88% chance of pulling three red marbles in a row.

Step-by-step explanation:

First pull = 4/15 (26.67%)

second pull = 3/14 (21.43%)

Third pull = 2/13 (15.38%)

You need to multiply these three fractions to get the probability of pulling three reds in a row, doing that will get you 4/455 or 0.88%

Answer:

z=10 i hope this will help you

Step-by-step explanation:

3z+5=35

3z=35-5

3z=30

z=10

Answer:

Z = 10

Step-by-step explanation:

3Z+5=35

Subtract 5 from both sides

3Z=30

Divide both sides by 3

Z=10

Answer:

[tex]=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]

Step-by-step explanation:

[tex]x^{\frac{1}{3}}\left(x^{\frac{1}{2}}+2x^2\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=x^{\frac{1}{3}},\:b=x^{\frac{1}{2}},\:c=2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+x^{\frac{1}{3}}\cdot \:2x^2\\=x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\\mathrm{Simplify}\:x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}:\quad x^{\frac{5}{6}}+2x^{\frac{7}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=x^{\frac{5}{6}}[/tex]

[tex]x^{\frac{1}{3}}x^{\frac{1}{2}}\\\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}\\x^{\frac{1}{3}}x^{\frac{1}{2}}=\:x^{\frac{1}{3}+\frac{1}{2}}\\=x^{\frac{1}{3}+\frac{1}{2}}\\\mathrm{Join}\:\frac{1}{3}+\frac{1}{2}:\quad \frac{5}{6}\\\frac{1}{3}+\frac{1}{2}\\\mathrm{Least\:Common\:Multiplier\:of\:}3,\:2:\quad 6\\Adjust\:Fractions\:based\:on\:the\:LCM\\=\frac{2}{6}+\frac{3}{6}[/tex]

[tex]\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}\\=\frac{2+3}{6}\\\mathrm{Add\:the\:numbers:}\:2+3=5\\=\frac{5}{6}\\=x^{\frac{5}{6}}\\2x^2x^{\frac{1}{3}}=2x^{\frac{7}{3}}\\=x^{\frac{5}{6}}+2x^{\frac{7}{3}}[/tex]

Answer:

£135 is the correct answer.

Step-by-step explanation:

Let C be the cost of table.

And let R be the radius of table.

Cost of table is directly proportional to square of radius.

As per question statement:

[tex]C\propto R^{2}[/tex] or

[tex]C=a\times R^2 ....... (1)[/tex]

where [tex]a[/tex] is the constant to remove the [tex]\propto sign[/tex].

It is given that

[tex]C_1 =[/tex] £60 and [tex]R_1 = 50\ cm[/tex]

[tex]C_2 = ?[/tex] when [tex]R_2= 75\ cm[/tex]

Putting the values of [tex]C_1[/tex] and [tex]R_1[/tex] in equation (1):

[tex]60=a \times 50^2 ....... (2)[/tex]

Putting the values of [tex]C_2[/tex] and [tex]R_2[/tex] in equation (1):

[tex]C_2=a \times 75^2 ....... (3)[/tex]

Dividing equation (2) by (3):

[tex]\dfrac{60}{C_2}= \dfrac{a \times 50^2}{a \times 75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{50^2}{75^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{2^2}{3^2}\\\Rightarrow \dfrac{60}{C_2}= \dfrac{4}{9}\\\Rightarrow C_2 = 15 \times 9 \\\Rightarrow C_2 = 135[/tex]

So, £135 is the correct answer.

Answer:the answer would be 4. Hope this helps.

Step-by-step explanation:

Using the formula of union of three events, the number of patrons who didn't like any of given coffee drinks = 4.

Union of three events : P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).

n (latte ∩ espressos) = 14

n (espressos ∩ cappuccinos) = 11

n (lattes ∩ cappuccinos) = 7

n (latte ∩ espressos ∩ cappuccinos) = 3

n (lattes) = 22

n (espressos) = 30

n (cappuccinos) = 23

n(latte ∪ espressos ∪ cappuccinos) =

= n (lattes) + n (espressos) + n (cappuccinos) - n (latte ∩ espressos) - n (espressos ∩ cappuccinos) - n (lattes ∩ cappuccinos) + n (latte ∩ espressos ∩ cappuccinos)

= 22 + 30 + 23 - 14 - 11 - 7 + 3

= 46

n (universe) = 50

Number of patrons who didn't like any of these drinks =  

= n (universe) - n (latte ∪ espressos ∪ cappuccinos) = 50 - 46 = 4

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Answer:

73.24% probability that 6 or more people from this sample are unemployed

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

The standard deviation of the binomial distribution is:

[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].

In this problem, we have that:

[tex]n = 100, p = 0.071[/tex]

So

[tex]\mu = E(X) = np = 10*0.071 = 7.1[/tex]

[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{100*0.071*0.929} = 2.5682[/tex]

What is the probability that 6 or more people from this sample are unemployed

Using continuity correction, this is [tex]P(X \geq 6 - 0.5) = P(X \geq 5.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 5.5. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{5.5 - 7.1}{2.5682}[/tex]

[tex]Z = -0.62[/tex]

[tex]Z = -0.62[/tex] has a pvalue of 0.2676

1 - 0.2676 = 0.7324

73.24% probability that 6 or more people from this sample are unemployed

Answer:

x = 6

Step-by-step explanation:

x + 6 = x + x

Combine like terms

x+6 =2x

Subtract x from each side

x+6-x = 2x-x

6 = x

Answer:

6163.2 years

Step-by-step explanation:

A_t=A_0e^{-kt}

Where

A_t=Amount of C 14 after “t” year

A_0= Initial Amount

t= No. of years

k=constant

In our problem we are given that A_t is 54% that is if A_0=1 , A_t=0.54

Also , k=0.0001

We have to find t=?

Let us substitute these values in the formula

0.54=1* e^{-0.0001t}

Taking log on both sides to the base 10 we get

log 0.54=log e^{-0.0001t}

-0.267606 = -0.0001t*log e

-0.267606 = -0.0001t*0.4342

t=\frac{-0.267606}{-0.0001*0.4342}

t=6163.20

t=6163.20 years

PLEASE MARK BRAINLY

Answer:

There were 210 downloads of the standard version.

Step-by-step explanation:

This question can be solved using a system of equations.

I am going to say that:

x is the number of downloads of the standard version.

y is the number of downloads of the high-quality version.

The size of the standard version is 2.6 megabytes (MB). The size of the high-quality version is 4.2 MB. The total size downloaded for the two versions was 4074 MB.

This means that:

[tex]2.6x + 4.2y = 4074[/tex]

Yesterday, the high-quality version was downloaded four times as often as the standard version.

This means that [tex]y = 4x[/tex]

How many downloads of the standard version were there?

This is x.

[tex]2.6x + 4.2y = 4074[/tex]

Since [tex]y = 4x[/tex]

[tex]2.6x + 4.2*4x = 4074[/tex]

[tex]19.4x = 4074[/tex]

[tex]x = \frac{4074}{19.4}[/tex]

[tex]x = 210[/tex]

There were 210 downloads of the standard version.

Answer:

The correct option is (d).

Step-by-step explanation:

The complete question is:

The random variable x represents the number of computers that families have along with the corresponding probabilities. Use the probability distribution table below to find the mean and standard deviation for the random variable x.

    x :    0          1          2           3         4

p (x) : 0.49     0.05    0.32     0.07    0.07

(a) The mean is 1.39 The standard deviation is 0.80

(b) The mean is 1.39 The standard deviation is 0.64

(c)The mean is 1.18 The standard deviation is 0.64

(d) The mean is 1.18 The standard deviation is 1.30

Solution:

The formula to compute the mean is:

[tex]\text{Mean}=\sum x\cdot p(x)[/tex]

Compute the mean as follows:

[tex]\text{Mean}=\sum x\cdot p(x)[/tex]

         [tex]=(0\times 0.49)+(1\times 0.05)+(2\times 0.32)+(3\times 0.07)+(4\times 0.07)\\\\=0+0.05+0.64+0.21+0.28\\\\=1.18[/tex]

The mean of the random variable x is 1.18.

The formula to compute variance is:

[tex]\text{Variance}=E(X^{2})-[E(X)]^{2}[/tex]

Compute the value of E (X²) as follows:

[tex]E(X^{2})=\sum x^{2}\cdot p(x)[/tex]

          [tex]=(0^{2}\times 0.49)+(1^{2}\times 0.05)+(2^{2}\times 0.32)+(3^{2}\times 0.07)+(4^{2}\times 0.07)\\\\=0+0.05+1.28+0.63+1.12\\\\=3.08[/tex]

Compute the variance as follows:

[tex]\text{Variance}=E(X^{2})-[E(X)]^{2}[/tex]

             [tex]=3.08-(1.18)^{2}\\\\=1.6876[/tex]

Then the standard deviation is:

[tex]\text{Standard deviation}=\sqrt{\text{Variance}}[/tex]

                              [tex]=\sqrt{1.6876}\\\\=1.2990766\\\\\approx 1.30[/tex]

Thus, the mean and standard deviation for the random variable x are 1.18 and 1.30 respectively.

The correct option is (d).