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3 years ago

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Alec has 2 comic books. Shereen has c times as many comic books as Alec. Write an expression that shows how many comic books She

reen has.
Type an asterisk ( * ) if you want to use a multiplication sign and a slash ( / ) if you want to use a division sign.

1 answer:

maks197457 [2]3 years ago

7 0

The expression is 2*c=b

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I have no clue what the answer is

spin [16.1K]

Answer:

Second option is the correct answer.

Step-by-step explanation:

$\frac{2}{3} x - 5 > 3 \\ \\ \frac{2}{3} x > 5 + 3\\ \\ \frac{2}{3} x > 8 \\ \\ x > 8 \times \frac{3}{2} \\ \\ x > 12$

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3 years ago

Find the shaded area. Use 3.14 as the approximate value of pi.

12345 [234]

Answer:

164.73 cm²

Step-by-step explanation:

Area of the shaded area = area of semi-circle - area of the triangle

Area of the semi-circle:

Area = ½(πr²)

r = ½ of 34 = 17 cm

π = 3.14

Area = ½(3.14 × 17²)

Area = 453.73 cm²

Area of the triangle:

Area = ½*b*h

b = 34 cm

h = 17 cm

Area = ½ × 34 × 17

Area = 289 cm²

Area of the shaded area = 453.73 - 289 = 164.73 cm²

3 0

3 years ago

What is true of the data in the dot plot? Check all that apply.

otez555 [7]

True answers for data in the plot are

The center is 13

The center is not 14

The Peak is 14.

It has three clusters

It is not symmetric to left or right it is bi-modal.

It is has a range from 10 to 15 most number of data points are 13 to 15.

Total number times Shelly waited is 16 times.

Step-by-step explanation:

While taking cumulative frequency 56.75 percentage comes in 13%.
It is the center point of the data.
The 14 is not the center as it shows 93rd percent.
It has three clusters 0-2 has one cluster,2-4 has 2nd cluster,5-8 third.
It is not skewed on the left is has bi modal frequency has two heights.
The person indeed waited for 16 times adding total dots.
There was a zero 12 which created bi-modal distribution

8 0

3 years ago

Read 2 more answers

Find the circumference of the circle with the given radius. Round to the nearest tenth. Use 3.14 for π.

Blababa [14]

16.3 cm that is the answer

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4 years ago

Read 2 more answers

The Aluminum Association reports that the average American uses 56.8 pounds of aluminum in a year. A random sample of 49 househo

KonstantinChe [14]

Answer:

a) 10.38% probability that the sample mean will be more than 59 pounds.

b) 67.72% probability that the sample mean will be more than 56 pounds.

c) 22.10% probability that the sample mean will be between 56 and 57 pounds.

d) 1.46% probability that the sample mean will be less than 53 pounds.

e) 0% probability that the sample mean will be less than 49 pounds.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

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

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

$Z = \frac{X - \mu}{\sigma}$

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.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

$\mu = 56.8, \sigma = 12.2, n = 49, s = \frac{12.2}{\sqrt{49}} = 1.74285$

a. More than 59 pounds

This is 1 subtracted by the pvalue of Z when X = 59. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{59 - 56.8}{1.74285}$

$Z = 1.26$

$Z = 1.26$ has a pvalue of 0.8962.

1 - 0.8962 = 0.1038

10.38% probability that the sample mean will be more than 59 pounds.

b. More than 56 pounds

This is 1 subtracted by the pvalue of Z when X = 56. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{56 - 56.8}{1.74285}$

$Z = -0.46$

$Z = -0.46$ has a pvalue of 0.3228.

1 - 0.3228 = 0.6772

67.72% probability that the sample mean will be more than 56 pounds.

c. Between 56 and 57 pounds

This is the pvalue of Z when X = 57 subtracted by the pvalue of Z when X = 56. So

X = 57

$Z = \frac{X - \mu}{s}$

$Z = \frac{57 - 56.8}{1.74285}$

$Z = 0.11$

$Z = 0.11$ has a pvalue of 0.5438

X = 56

$Z = \frac{X - \mu}{s}$

$Z = \frac{56 - 56.8}{1.74285}$

$Z = -0.46$

$Z = -0.46$ has a pvalue of 0.3228.

0.5438 - 0.3228 = 0.2210

22.10% probability that the sample mean will be between 56 and 57 pounds.

d. Less than 53 pounds

This is the pvalue of Z when X = 53.

$Z = \frac{X - \mu}{s}$

$Z = \frac{53 - 56.8}{1.74285}$

$Z = -2.18$

$Z = -2.18$ has a pvalue of 0.0146

1.46% probability that the sample mean will be less than 53 pounds.

e. Less than 49 pounds

This is the pvalue of Z when X = 49.

$Z = \frac{X - \mu}{s}$

$Z = \frac{49 - 56.8}{1.74285}$

$Z = -4.48$

$Z = -4.48$ has a pvalue of 0.

0% probability that the sample mean will be less than 49 pounds.

7 0

3 years ago