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3 years ago

HEY CAN ANYONE PLS ANSWER DIS MATH QUESTION!!!!

Mathematics

1 answer:

In-s [12.5K]3 years ago

4 0

Answer:

I believe the answer could be c

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Which of the following is an offer? Select one: a. Hedva walks into the Ferrari Dealership and says, “I am going to buy a car.”

rewona [7]

Answer:

D, "I will pay $25 for this book."

Step-by-step explanation

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4 years ago

Read 2 more answers

Choose all numbers divisible by 2

Ivahew [28]

Answer:

all real numbers

Step-by-step explanation:

5 0

3 years ago

I have no clue how to do this pls help this is for 15 points and ill do a brainliest aswell

Harman [31]

Answer:

you can!!

Step-by-step explanation:

you can break this apart into whole numbers and its fractional part.

-20+-12= -32

-3/4 + -1.4 = -1

-32 + -1

=-33

3 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

What is 2/8 in simpilist form

Tems11 [23]

Answer:

1/4

Step-by-step explanation:

4 0

3 years ago