6x3 + 8x2 – 2x + 4 and 10x3 + x2 + 11x + 9

Cindy went to the salon to get highlights and a haircut. Her total cost was $95. Cindy decided to tip her stylist 22% because sh

yan [13]

Her tip would be $20.90

8 0

3 years ago

A football stadium has 2,500 seats. For the first game of the season, 2,075 seats were filled. Which percent of seats were fille

malfutka [58]

83 % of the seats

2075 / 2500 x 100

5 0

3 years ago

Read 2 more answers

The ammunition storage room has 10 feet between the floor and the ceiling. Each box of ammunition is 10 feet tall. Each crate of

KIM [24]

Answer:

the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches. SO the answer is that a maximum of 10 crates can be stacked from floor to ceiling.

Step-by-step explanation:

i) the maximum number of crates that can be stacked between the floor and ceiling

$\frac{height \hspace{0.1cm} of \hspace{0.1cm} ammunition \hspace{0.1cm} box}{height \hspace{0.1cm} of \hspace{0.1cm} each \hspace{0.1cm} crate} = \frac{10 \hspace{0.1cm} feet}{12 \hspace{0.1cm} inches} = \frac{10 \times 12 \hspace{0.1cm} inches }{12 \hspace{0.1cm} inches} = 10 \hspace{0.1cm} crates$

where 1 foot = 12 inches

3 0

3 years ago

A college entrance exam company determined that a score of 2323 on the mathematics portion of the exam suggests that a student i

aleksandrvk [35]

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu=23\\\\H_a:\mu> 23$

c) Test statistic t=1.53

P-value=0.064

d) The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that students who complete the core curriculum are ready for college-level mathematics. That is that the true score for the group is not significantly higher than 23.

Step-by-step explanation:

The question is incomplete:

a) State the appropriate null and alternative hypotheses.

c) Use the P-value approach at the 0.05 level of significance to test the hypotheses in part (a). ldentify the test statistic. (Round to two decimal places as needed.) Identfy the P-value. P-value (Round to three decimal places as needed.)

d) Write a conclusion based on the results. Choose the correct answer below. ? the null hypothesis and claim that there ? sufficient evidence to conclude that the population mean is ? than 20.

This is a hypothesis test for the population mean.

The claim is that students who complete the core curriculum are ready for college-level mathematics.

Then, the null and alternative hypothesis are:

$H_0: \mu=23\\\\H_a:\mu> 23$

The significance level is assumed to be 0.05.

The sample has a size n=150.

The sample mean is M=23.4.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=3.2.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{3.2}{\sqrt{150}}=0.261$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{23.4-23}{0.261}=\dfrac{0.4}{0.261}=1.531$

The degrees of freedom for this sample size are:

$df=n-1=150-1=149$

This test is a right-tailed test, with 149 degrees of freedom and t=1.531, so the P-value for this test is calculated as (using a t-table):

$P-value=P(t>1.531)=0.064$

As the P-value (0.064) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that students who complete the core curriculum are ready for college-level mathematics.

6 0

3 years ago

Please help. I’ll mark you as brainliest if correct!

In-s [12.5K]

Answer:

8

Step-by-step explanation:

f^-1(x) =dy/dx

dy/dx(8x+9)=8

8 0

3 years ago