All categories

4 years ago

Please can someone help me?

Mathematics

2 answers:

Allushta [10]4 years ago

8 0

The answer is 53, since it’s not green, it’s greater than 50, and it has a 3 in the ones place

Inga [223]4 years ago

4 0

The special number is 53 in the purple circle i know this because from Jeffrey's clues hes says its not green it has three in the ones place 5(3) <----ones place and is over 50 the only one to fit this category is 53 purple circle
<span />

You might be interested in

An employer interviews 12 people for four openings at a company. Five of the 12 people are women. All 12 applicants are qualifie

MA_775_DIABLO [31]

Answer:

a) The employer can fill the four positions in 495 ways.

b) The employer can fill the four positions in 210 ways.

Step-by-step explanation:

The order is not important.

For example:

Selecting John, Laura, Mary and Tre'Davious is the same as selecting Laura, John, Mary and Tre'Davious.

So we use the combinations formula to solve this problem.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

(a) the selection is random

There are 12 people, and 4 are selected. So

$C_{12,4} = \frac{12!}{4!8!} = 495$

(b) exactly two selections are women?

There are 7 men and 5 women. We want to select 2 men and 2 women. So

$C_{7,2}*C_{5,2} = \frac{7!}{2!5!}*\frac{5!}{2!3!} = 210$

3 0

4 years ago

V : 33= 7 : 11 what is the answer to this please

UkoKoshka [18]

The value of V illustrated in the ratio V : 33= 7 : 11 is 21.

<h3>What is a ratio?</h3>

It should be noted that a ratio is simply used to show the relationship between the variables and illustrate the comparison.

In this case, the value of V can be illustrated as:.

V/33 = 7/11

Cross multiply

11V = 33 × 7

V = (33 × 7) / 11

V = 21

Therefore, V is 21.

Learn more about ratio on:

brainly.com/question/2328454

#SPJ1

4 0

1 year ago

The scatterplot contains data points, including the

ser-zykov [4K]

Step-by-step explanation:

what is the question? please provide a question.

5 0

3 years ago

Your friend is graphing the point (2,5) 2 , 5 . Her first step is to start at the origin and go up 2 2 units. Is her method corr

Alchen [17]

No. You need to first move along the x-axis.

3 0

3 years ago

Read 2 more answers

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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

$Z = \frac{670 - 660}{9}$

$Z = 1.11$

$Z = 1.11$ has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0

2 years ago