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VladimirAG [237]
3 years ago
11

Two positive consecutive odd numbers of 143

Mathematics
2 answers:
NNADVOKAT [17]3 years ago
5 0
You can't have two odd numbers equal an odd number(143).
goldfiish [28.3K]3 years ago
5 0
13 * 11 = 143

Hope that helps!
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What is the inverse of the function f(x) = x – 12?
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y=x-12\\
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3 years ago
Al, Tom and Joe share €3000.
agasfer [191]

Answer:

Tom's share = €800

Step-by-step explanation:

Given that:

Amount shared by Al, Tom and Joe = €3000

Ratio of Al's share to Tom's share = 5 : 4

Joe's share = 1.5(Tom's share) = 1.5(4) = 6

Therefore,

Ratio of Al, Tom and Joe = 5 : 4 : 6

Let,

x be the number of times of amount each get.

5x + 4x + 6x = 3000

15x = 3000

Dividing both sides by 15

\frac{15x}{15}=\frac{3000}{15}\\x=200

Tom's share = 4x = 4(200) = €800

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Tom's share = €800

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3 years ago
100,000 equals 10 to the what power?
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100,000 equals 10 to the power of 5, hope helps.
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3 years ago
Read 2 more answers
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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3 years ago
True or False: if the product of three numbers is negative, then all the numbers are negative?
S_A_V [24]
This question would be false
3 0
3 years ago
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