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Ket [755]
3 years ago
10

Peter makes $7000 a month plus some money by commission rates. He gets 6% of everything he sells. If peter sold $55000 worth of

items this month, what is his salary for the month?
Mathematics
1 answer:
tatiyna3 years ago
7 0
He made a total of $10,300 that month
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Step-by-step explanation:

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Will mark the correct answer brainliest
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How long the ball was in the air

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3.5 seconds is the amount of time that the ball's height starts from 0 meters and ends to 0 meters.

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Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abrade
WITCHER [35]

Answer:

The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

H_0: \mu_d=0\\\\H_a: \mu_d>0

We have n=8 pairs of data. We calculate the difference as:

d_1=U_1-A_1=36.4-28.5=7.9

Then, with this procedure we get the sample for d:

d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]

The sample mean and standard deviation are:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729

The degrees of freedom for this sample size are:

df=n-1=8-1=7

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>1.729)=0.0637

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

If we use the critical value approach, the critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998. The test statistic is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

8 0
3 years ago
Ten college students were randomly selected. Their grade point averages​ (GPAs) when they entered the program were between 3.5 a
solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

\hat y=\alpha +\hat \beta x

The (1 - <em>α</em>)% confidence interval for the regression coefficient or slope (\hat \beta ) is:

Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

\hat Y=3.584756+0.090953 X

The summary of the regression analysis is:

Predictor          Coefficient             SE             t-stat            p-value

Constant             3.584756          0.078183       45.85075      5.66 x 10⁻¹¹

Entering GPA   0.090953          0.022162        4.103932       0.003419

The regression coefficient and standard error are:

\hat \beta = 0.090953\\SE (\hat \beta)=0.022162

The critical value of <em>t</em>  for 95% confidence level and 8 degrees of freedom is:

t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306

Compute the 95% confidence interval for (\hat \beta ) as follows:

CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)

Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).

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3 years ago
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Scrat [10]
Matthew will walk 3 \frac{1}{8} miles to the bus stop in 5 days.
5 0
2 years ago
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