Ask question

Ask question

All categories

3 years ago

14

On friday night 165 people saw the dinosaur exhibit at the natural history museum. this amount represents 22% of the people who

visited the museum that night. select from the drop-down menu to correctly identify the total number of people who visited the museum friday night

1 answer:

olga nikolaevna [1]3 years ago

7 0

I think the answer is 36. hope it helps

You might be interested in

Micheal is taking a survey of students at his high school to find out how many hours they work per week. He surveys all of the s

yulyashka [42]

No because that would just be measuring freshman a better sampling method would be surveying the same number of students for each grade

6 0

3 years ago

Evaluate |2 5/6 +y| for y= 7/4

bearhunter [10]

Answer:

Step-by-step explanation:

7 0

2 years ago

Select the correct answer. Simplify the expression.<br> (see the screenshot)

Sliva [168]

Answer:

Option D

Step-by-step explanation:

Given expression has been given as,

$\sqrt[5]{224x^{11}y^8}$

$\sqrt[5]{224x^{11}y^8}=\sqrt[5]{2\times 2\times 2\times 2\times 2\times 7(x^{11})(y^8)}$

$=\sqrt[5]{(2^5)\times (7)(x^{10}\times x)(y^5\times y^3)}$

$=2^{\frac{5}{5}}\times 7^{\frac{1}{5}}\times x^{\frac{10}{5}}\times x^{\frac{1}{5}}\times y^{\frac{5}{5} }\times y^{\frac{3}{5} }$

$=2\times 7^{\frac{1}{5}}\times x^2\times y\times x^{\frac{1}{5} }\times y^{\frac{3}{5} }$

$=2x^2y\sqrt[5]{7xy^3}$

Option D will be the answer.

7 0

3 years ago

Professor Halen teaches a College Mathematics class. The scores on the midterm exam are normally distributed with a mean of 72.3

lbvjy [14]

Answer:

14.63% probability that a student scores between 82 and 90

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 72.3, \sigma = 8.9$

What is the probability that a student scores between 82 and 90?

This is the pvalue of Z when X = 90 subtracted by the pvalue of Z when X = 82. So

X = 90

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

$Z = \frac{90 - 73.9}{8.9}$

$Z = 1.81$

$Z = 1.81$ has a pvalue of 0.9649

X = 82

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

$Z = \frac{82 - 73.9}{8.9}$

$Z = 0.91$

$Z = 0.91$ has a pvalue of 0.8186

0.9649 - 0.8186 = 0.1463

14.63% probability that a student scores between 82 and 90

3 0

3 years ago

What is the value of x? 140 34 56

olga55 [171]

I don’t know what you’re tying to say

3 0

2 years ago