Ask question

Ask question

All categories

3 years ago

8

Michael has 4 red shirts, 3 green shirts, and 5 blue shirts in his closet. If he randomly selects a shirt what is the probabilit

y that he chooses blue or red?

1 answer:

Sedaia [141]3 years ago

4 0

Answer:

3/4

Step-by-step explanation:

You might be interested in

What is the solution to this inequality?

Evgen [1.6K]

Answer:

Answer: x<7.1

Step-by-step explanation:

${ \tt{4.7 > x - 2 \frac{2}{5} }} \\ \\ { \tt{4.7 > x - \frac{12}{5} }} \\ \\ { \tt{x < 4.7 + \frac{12}{5} }} \\ \\ { \boxed{ \tt{x < 7.1}}}$

6 0

2 years ago

lily sees 15 tan puppies and 8 white puppies at the pet store.how many more tan puppies than white puppies does she sees? Draw p

Nataliya [291]

Answer:7

Step-by-step explanation:

Just put 7 as your answe

7 0

3 years ago

Read 2 more answers

Janet is painting her shed. She buys some gallon-sized paint cans that cost $15 each and 2 quart-sized paint cans that cost $4 e

blondinia [14]

1 gallon cost $15

1 quarts cost $4

1 quarts = 1/4 gallon

15x for number of gallon cans

2 cans at $4 each = $8

15x + 8 = 53

Subtract 8 from both sides:

15x = 53-8

15x = 45

Divide both sides by 15:

x = 45/15

x = 3

She bought 3 gallon cans of paint

Plus she bought 2 quart cans, which is equal to a half gallon.

She bought a total of 3.5 gallons of paint.

7 0

3 years ago

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its average number of unoc

Bingel [31]

Answer:

We need a sample size of at least 719

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large a sample size is required to vary population mean within 0.30 seat of the sample mean with 95% confidence interval?

This is at least n, in which n is found when $M = 0.3, \sigma = 4.103$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.3 = 1.96*\frac{4.103}{\sqrt{n}}$

$0.3\sqrt{n} = 1.96*4.103$

$\sqrt{n} = \frac{1.96*4.103}{0.3}$

$(\sqrt{n})^{2} = (\frac{1.96*4.103}{0.3})^{2}$

$n = 718.57$

Rouding up

We need a sample size of at least 719

6 0

3 years ago

I need help with this question

Julli [10]

Answer:

it's 324

Step-by-step explanation:

divide 72 by 4 to get one side, then multiply the side by itself, 18 times 18

3 0

3 years ago