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3 years ago

The coordinates of the endpoints of CD

1 answer:

7 0

Use the first part of the midpoint formula:

(x1 + x2)/2

Plug in what we know:

(-4 + 14)/2

10/2

5

So the x-value is 5.

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Robyn opened a bank account to save her birthday money. it was paying 3.75% interest. then the interest rate went up by 0.65%. h

Savatey [412]

Her new interest rate would be 4.40%

8 0

3 years ago

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A survey among freshman at a certain university revealed that the number of hours spent studying before final exams was normally

patriot [66]

Answer: 0.0775

Step-by-step explanation:

Given : Mean : $\mu = 25$

Standard deviation : $\sigma =15$

Sample size : $n=36$

Since its normal distribution , then the formula to calculate the z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= 28.2 hours

$z=\dfrac{28.2-25}{\dfrac{15}{\sqrt{36}}}=1.28$

For x= 30 hours

$z=\dfrac{30-25}{\dfrac{15}{\sqrt{36}}}=2$

The P- value = $P(1.28$

$=P(z$

Hence, the probabiliy that the average time spent stydying for the sampe was between 28.2 and 30 hours = 0.0775

6 0

3 years ago

I don't know something about a shape blah blah blahhhhhh

PolarNik [594]

Add all the numbers on the shape until you get your answer

5 0

2 years ago

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri

nlexa [21]

Answer:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday Tuesday Wednesday Thursday Friday Saturday Total

12 9 11 10 9 9 60

The level of significance assumed for this case is $\alpha=0.05$

The statistic to check the hypothesis is given by:

$\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{60}{6}= 10$ and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

$\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8$

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

$df=categor-1=6-1=5$

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got $\chi^2_{critc}= 11.0705$

a. 11.070

And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0

2 years ago

Differentiate 3x^2-5x

weqwewe [10]

Answer:

6x - 5

Step-by-step explanation:

$Let \: y = 3 {x}^{2} - 5x \\ \\ \frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2} - 5x) \\ \\ = 3 \frac{d}{dx} {x}^{2} - 5 \frac{d}{dx} x \\ \\ = 3(2x) - 5(1) \\ \\ \boxed{\boxed{\huge \purple{\frac{dy}{dx}} \pink{ = } \orange{6x - 5}}}$

3 0

2 years ago