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3 years ago

Help me please it’s due soon

Mathematics

1 answer:

prisoha [69]3 years ago

8 0

The solution is negative.

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Divide. Express your answer in simplest form.<br><br> 10 ÷ 3

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3.33333 , it’s continuous

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3 years ago

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Jake studied x hours for a big test. Julie studied half as long. Write an algebraic expression for long julie studied.

kipiarov [429]

X/2 is howling Julie studied

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4 years ago

W(x)= -2x -2 what does w(9)=

natka813 [3]

Answer:

w(9) = -20

Step-by-step explanation:

In order to find this, take the w function and input 9 in for all the x values.

w(x) = -2x - 2

w(9) = -2(9) - 2

w(9) = -18 - 2

w(9) = -20

5 0

3 years ago

Need help please answer

fomenos

Answer:62.73

sin-1 8 divide by 9

6 0

3 years ago

A person drilled a hole in a die and filled it with a lead weight, then proceeded to roll it 200 times. Here are the observed f

Olin [163]

Answer:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference in the frequencies

H1: There is a difference in the frequencies

The level of significance assumed for this case is $\alpha=0.01$

The statistic to check the hypothesis is given by:

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

The observed values are:

Value 1 2 3 4 5 6

Frequency 26 32 44 37 27 34

And the expected values are for this case the same $E_i = \frac{200}{6}= 33.33$

And now we can calculate the statistic:

$\chi^2 = \frac{(26-33.33)^2}{33.33}+\frac{(32-33.33)^2}{33.33}+\frac{(44-33.33)^2}{33.33}+\frac{(37-33.33)^2}{33.33}+\frac{(27-33.33)^2}{33.33}+\frac{(34-33.33)^2}{33.33}=6.7$

Now we can calculate the degrees of freedom for the statistic given by:

$df=6-1=5$

And we can calculate the p value given by:

$p_v = P(\chi^2_{5} >6.7)=0.244$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis so then we can conclude that the outcomes are equally likely

7 0

4 years ago