H=-16(t-) + 49,<br> Does the marshmallow reach its maximum height?

Trevor plans on buying a car. Four banks offer the annual interest rates shown.

mote1985 [20]

Solution

In order to find greatest to least number, we have to convert the given fractions into decimals.

3 $\frac{1}{10}$ % = 3 + 1/10 = 3 + 0.10 = 3.10%

3 $\frac{2}{5}$ % = 3 + 2/5 = 3 + 0.40 = 3.40%

3 $\frac{1}{4}$ %= 3 + 0.25 = 3.25%

3 $\frac{1}{3}$ % = 3 + 0.33 = 3.33%

Now we can easily, compare the numbers and write from greatest to least.

3.40%, 3.33%, 3.25%, 3.10%

Greatest to least = 3 2/5%, 3 1/3%3, 3 1/4%, 1/10%,

Thank you :)

8 0

2 years ago

Read 2 more answers

Wanda jumped rope 25 times every minute. What is the algebraic equation to express the function of the total number of times Wan

jasenka [17]

The answer is a m=25j

8 0

2 years ago

Does the line y = 3x + 1 pass through the point (4,15)

Firdavs [7]

No. when the line reaches 4 on the x axis it is at 13 on the y axis.

3 0

1 year ago

In a small town, the stray cat population is rapidly increasing. There are currently 20 stray cats, and it is estimated

BigorU [14]

2 years will be 80 cats (20 x 2 = 40 (year 1) and 40 x 2 = 80 (year 2))

3 years will be 160 (80 x 2 = 160)

5 years will be (160 x 2 = 320 (year 4) and 320 x 2 = 640 (year 5)) So 5 years will be 640

7 0

3 years ago

A company wants to find out if the average response time to a request differs across its two servers. Say µ1 is the true mean/ex

lorasvet [3.4K]

Answer:

a) The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

Test statistic t=0.88

The P-value is obtained from a t-table, taking into acount the degrees of freedom (419) and the type of test (two-tailed).

b) A P-value close to 1 means that a sample result have a high probability to be obtained due to chance, given that the null hypothesis is true. It means that there is little evidence in favor of the alternative hypothesis.

c) The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

d) The consequences of the confidence interval containing 0 means that the hypothesis that there is no difference between the response time (d=0) is not a unprobable value for the true difference.

This relate to the previous conclusion as there is not enough evidence to support that there is significant difference between the response time, as the hypothesis that there is no difference is not an unusual value for the true difference.

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that there is significant difference in the time response for the two servers.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0$

The significance level is 0.05.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

$M_d=M_1-M_2=12.5-12.2=0.3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{3^2}{196}+\dfrac{4^2}{225}}\\\\\\s_{M_d}=\sqrt{0.046+0.071}=\sqrt{0.117}=0.342$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.3-0}{0.342}=\dfrac{0.3}{0.342}=0.88$

The degrees of freedom for this test are:

$df=n_1+n_2-1=196+225-2=419$

This test is a two-tailed test, with 419 degrees of freedom and t=0.88, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>0.88)=0.381$

As the P-value (0.381) is greater 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 there is significant difference in the time response for the two servers.

<u>Confidence interval </u>

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=196 has a mean of 12.5 and a standard deviation of 3.

The sample 2, of size n2=225 has a mean of 12.2 and a standard deviation of 4.

The difference between sample means is Md=0.3.

The estimated standard error of the difference is s_Md=0.342.

The critical t-value for a 95% confidence interval and 419 degrees of freedom is t=1.966.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_{M_d}=1.966 \cdot 0.342=0.672$

Then, the lower and upper bounds of the confidence interval are:

$LL=M_d-t \cdot s_{M_d} = 0.3-0.672=-0.372\\\\UL=M_d+t \cdot s_{M_d} = 0.3+0.672=0.972$

The 95% confidence interval for the difference in the two servers population expectations is (-0.372, 0.972).

7 0

2 years ago

H=-16(t-) + 49, Does the marshmallow reach its maximum height?