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2 months ago

How many pairs of Whole numbers have a sum of 40

2 answers:

6 0

Whole numbers include counting numbers starting from zero {0, 1, 2, 3, 4, ...} and are all positive.
If pairs of whole numbers sum to 40, the total number of pairs is 41 since each time you decrease one number by 1 and increase the other by 1, the sum remains 40.

Examples:
40 + 0 = 40
39 + 1 = 40
38 + 2 = 40
...

lawyer [12.5K]2 months ago

5 0

Prime numbers up to 40 are:

2, 3, 5, 7, 11, 13, 17, 23, 29, 31, 37.

There exist three prime pairs whose sums equal 40: {3, 37}, {11, 29}, and {17, 23}.

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Explain what it means to have a correlation in a scatterplot.

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Response:

A correlation is present in a scatter plot when a clear trend is observed in the outputs as the inputs rise. A general increase in the outputs signals a positive correlation, while a general decrease indicates a negative correlation.

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Which table contains only corresponding x-values and y-values where the value of y is one more than the product of x and 2?

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Answer:

Option (C)

Step-by-step explanation:

The value of y exceeds the multiplication of x by 2.

The equation that describes this condition is:

y = 2x + 1

Thus, by inserting x values into this equation, we can produce a table reflecting the corresponding input and output values:

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1 month ago

The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 20

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Answer:

a) Null and alternative hypothesis

$H_0: \mu=1503\\\\H_a:\mu< 1503$

b) Point estimate d = -$78

c) Test statistic t = -2.438

P-value = 0.0113

Reject H0. This indicates that the average automobile premium in Pennsylvania is lower than in the nation.

Step-by-step breakdown:

This is a statistical test for the average population mean.

The hypothesis posits that car insurance in Pennsylvania is notably less expensive compared to the national average.

Accordingly, the null and alternative hypotheses are:

$H_0: \mu=1503\\\\H_a:\mu< 1503$

The significance level is set at 0.05.

The sample size is n=25.

The sample mean equates to M=1425.

A point estimate of the difference between the Pennsylvania mean premium and the national average can be computed using the sample mean:

$d=M-\mu=1425-1503=-78$

Given that the standard deviation of the population is unknown, we approximate it using the sample standard deviation, which is s=160.

The estimated standard error of the mean is determined with the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{160}{\sqrt{25}}=32$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1425-1503}{32}=\dfrac{-78}{32}=-2.438$

The degrees of freedom for this sample size stand at:

$df=n-1=25-1=24$

This constitutes a left-tailed test, with 24 degrees of freedom and t=-2.438, rendering the P-value as (per a t-table):

$\text{P-value}=P(t$

As the P-value (0.0113) falls below the significance level (0.05), the results prove significant.

Thus, the null hypothesis obtains dismissal.

At a 0.05 significance level, there's sufficient evidence to assert that car insurance in Pennsylvania costs notably less than the national average.

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1 month ago

In a sample of 88 adults selected randomly from one town, it is found that 6 of them have been exposed to a particular strain of

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Response with clarification:

Let p denote the proportion of adults in the town who have encountered this flu strain.

According to the provided information

$H_0:p=0.08\\\\ H_a: p\neq0.08$

∵ $H_a$ this is a two-tailed test.

Test statistic:

$z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}$

, where p= denotes the population proportion

${\hat{p}$ = signifies the sample proportion

n= represents the sample size

Setting n= 6 and ${\hat{p}=\dfrac{6}{88}\approx0.068$ and p=0.08

$z=\dfrac{0.068-0.08}{\sqrt{\dfrac{0.08(1-0.08)}{88}}}$

$z=\dfrac{-0.012}{0.0289199522192}\approx-0.415$

P-value for the two-tailed test:[2P(Z>|z|)

=2P(Z>|-0.415|)

=2P(Z>0.415) = 2[1-P(Z≤0.415)] [∵ P(Z>z)=1-P(Z≤z)]

=2(1-0.6609) [from the z-table]

=0.6782

Decision: Because the p-value(0.6782) exceeds the significance level of 0.01, we do not reject the null hypothesis.

This leads us to conclude that there is insufficient evidence to back the assertion that the percentage of all adults in this town exposed to this flu strain deviates from the national average of 8%.

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