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3 years ago

Use the cubic model y=3x^3 to find the value of y when x=8

Mathematics

2 answers:

GalinKa [24]3 years ago

7 0

Answer:

Value of y is 1536.

Step-by-step explanation:

We have given an expression $y=3x^3$

We need to find the value of y when x=8

We will substitute the value of x as 8 in the given expression we get

$y=3(8)^3$

On simplification we get

$y=3(512)$

On further simplification we get:

$y=1536$

Hence, the required value that is the value of y is 1536

SpyIntel [72]3 years ago

6 0

Pemdas

exponent first

y=3(8³)
8³=512
y=3(512)
y=1536

the value is 1536

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It cost Hudson $28.20 to send 188 text messages. How much would it cost to send 63 text messages? Good luck!

frosja888 [35]

Answer:

$9.45

Step-by-step explanation:

1. First, you want to figure out how much one text message costs. To do this, divide the dollar amount (28.20) by the number of text messages (188). You get 0.15, or $0.15 per text message.

2. Using this $0.15 per text message, you can calculate how much 63 text messages costs. You simply multiply $0.15 by 63 to get $9.45, which is your answer.

Hope this helps! :)

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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).

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