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3 years ago

Fill in the table using this function rule. y=-3x + 5

Mathematics

1 answer:

zhannawk [14.2K]3 years ago

8 0

Answer:

y=-3(-4)+5,,,, you get y=12+5 y=17

y=-3(-2)+5,,,, you get y=6+5 y=11

y=-3(0)+5,,,, you get y=0+5 y=5

y=-3(2)+5,,,, you get y=-6+5 y=-1

Step-by-step explanation:

<h2><u><em>hope it helps:)</em></u></h2>

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Geometric mean between 2 and 24

svetoff [14.1K]

Answer:

6.9282

step by step:

=( 2 x 24 )1/2
= ( 48 )0.5
= 6.9282

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2 years ago

Jordan went to shelter to adopt a dog. There were 10 brown dogs and 10 black dogs. Jordan picked a dog, went home, then decided

ivanzaharov [21]

B. 9/19

Step-by-step explanation:

there was 20 dogs in total he took 1 black dog now 19 in total, now 9 black dogs and 10 brown dogs so one less probability of black dogs so 9 black dogs and 10 brown dogs so 9/19

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3 years ago

To amend a country's constitution, 5/6 of the 44 states in that country must approve the amendment.If 35 states approve an amend

Reptile [31]

No because 35/44 is like 79.3% and 5/6 is like 83%

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3 years ago

Which expression is equivalent to 5^3?

AleksAgata [21]

Answer:

C and D

Step-by-step explanation:

5^3 - 5^0 = 125 - 1 = 124, so it's not A

5^12 / 5^4 = 5^(12-4) = 5^8, so it's not B

5^7 * 5^-4 = 5^(7+(-4)) = 5^3, so it can be C

5^0 * 5^3 = 5^(0+3) = 5^3, so it can be D

5 + 5^2 = 5 + 25 = 30, so it can't be E

3 0

2 years ago

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon

fiasKO [112]

Answer:

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

A store randomly samples 603 shoppers over the course of a year and finds that 142 of them made their visit because of a coupon they'd received in the mail.

This means that $n = 603, \pi = \frac{142}{603} = 0.2355$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 - 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2016$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2355 + 1.96\sqrt{\frac{0.2355*0.7645}{603}} = 0.2694$

The 95% confidence interval for the fraction of all shoppers during the year whose visit was because of a coupon they'd received in the mail is (0.2016, 0.2694).

8 0

2 years ago