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3 years ago

I NEED THE ANSWER ASAP PLEASEE

Mathematics

2 answers:

Karolina [17]3 years ago

6 0

Answer:

Step-by-step explanation:

30 •0.65 = 19.5

Sam cannot make half a basket so he makes 19

AveGali [126]3 years ago

6 0

I believe it’s 19 let me know if I’m wrong

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Joseph needs 345 cement blocks to build a wall 40 cm thick. He needs 205 cement blocks to build a wall 25 cm thick. Each block c

IrinaVladis [17]

9514 1404 393

Answer:

$105

Step-by-step explanation:

The thicker wall takes 345 -205 = 140 more blocks. Each of those costs $0.75, so the added blocks will cost ...

140 × $0.75 = $105

The thicker wall will cost $105 more.

7 0

3 years ago

Solve the following linear equation for k.<br> 2 (k-5) + 3k = k + 6

kiruha [24]

Answer:

2(k-5)+3k=k+6

-3k. -3k

2(k-5)=-2k+6

/2. /2

k-5=-k+3

+k. +k

2k-5=3

+5. +5

2k=8

/2. /2

k=4

7 0

3 years ago

If the smallest of three consecutive numbers is k, write the expression for their sum.

omeli [17]

If they are three consecutive integers then they are:

k, k+1, and k+2 and the sum would then be:

s=k+k+1+k+2

s=3k+3

5 0

3 years ago

Without subtracting 8.5 - 4.64, determine what digit will be in the hundredths place. Explain

murzikaleks [220]

8.5 is the same as 8.50, right? So just take the digit in the hundredth place of both decimals and subtract. You'd be doing 0-4 so you can go ahead and borrow a number to make 10-4, which equals 6. So without subtracting 8.5 and 4.64, you can determine that 6 will be in the hundredth place. Hope this helps!

3 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

3 years ago