F(x) = –3x² - 6x +1<br><br> Find f(8)

1) Jose saves $9/ week.He wants to buy a backpack that cost $45. If he save his money for 7 weeks, Would he be able to afford th

Alja [10]

1. Yes he would be able to but the back pack and he would actually have 18 to spare. Explanation- he saves $9 a week so if he saves for 7 weeks then he will have $63 but since he only needed 45 he will have 18 to spare
2. Since they have 65.50 but need to have 180 then u could just subtract 180 from 65.50 so they would 114.5 more dollars
3. Since lee scored 4 FEWER this tells us that micheal has more points so you would just add 4 to micheals total of points which would give you that micheal had 22 points

3 0

3 years ago

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Which of the following states that if the product of two real numbers is zero,then either of the two is equal to zero or both nu

Lesechka [4]

Answer:

Zero Product Property

Step-by-step explanation:

If you have the following options:

A. Multiplication property

<u>B. Zero product property</u>

C. Identity property

D. Transitive property

3 0

3 years ago

Select all expressions shown that are equivalent to

Murrr4er [49]

Answer:

A. and C.

if you simplify the expressions, you will see which ones are equal to the expression given

4 0

2 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 30 people with 16 hav

Furkat [3]

Answer:

There is no significant evidence that p1 is different than p2 at 0.01 significance level.

99% confidence interval for p1-p2 is -0.171 ±0.237 that is (−0.408, 0.066)

Step-by-step explanation:

Let p1 be the proportion of the common attribute in population1

And p2 be the proportion of the same common attribute in population2

$H_{0}$ : p1-p2=0

$H_{a}$ : p1-p2≠0

Test statistic can be found using the equation:

$z=\frac{p2-p1}{\sqrt{{p*(1-p)*(\frac{1}{n1} +\frac{1}{n2}) }}}$ where

p1 is the sample proportion of the common attribute in population1 ( $\frac{16}{30} =0.533$ )

p2 is the sample proportion of the common attribute in population2 ( $\frac{1337}{1900} =0.704$ )

p is the pool proportion of p1 and p2 ( $\frac{16+1337}{30+1900}=0.701$ )

n1 is the sample size of the people from population1 (30)

n2 is the sample size of the people from population2 (1900)

Then $z=\frac{0.704-0.533}{\sqrt{{0.701*0.299*(\frac{1}{30} +\frac{1}{1900}) }}}$ ≈ 2.03

p-value of the test statistic is 0.042>0.01, therefore we fail to reject the null hypothesis. There is no significant evidence that p1 is different than p2.

99% confidence interval estimate for p1-p2 can be calculated using the equation

p1-p2± $z*\sqrt{\frac{p1*(1-p1)}{n1}+\frac{p2*(1-p2)}{n2}}$ where