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3 years ago

Please Help me on this one Is for tomorrow and I don’t understand

Mathematics

1 answer:

DerKrebs [107]3 years ago

8 0

For the first triangle add 45 and 95 and subtract it from 180

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Find the surface Area.

kompoz [17]

Answer: 20

Step-by-step explanation: how i got the answer was to multiply everything by two. then add everything together.

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3 years ago

Why do we say that a number is squared when raised to the power of 2

yawa3891 [41]

Answer:

We say that it is squared because we are multiplying that number twice. A square has two sides, therefore we are squaring the number.

I hope this helps. Please mark me Branliest if It did! Thank you and have a nice day!

4 0

3 years ago

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Bruce flipped a penny 5 times. It landed on heads 2 times and tails 3 times. If he flips the coin a 6th time, what statement is

babunello [35]

Answer:

Step-by-step explanation:

I would say B because they're only 2 options head and tails.

✩brainliest appreciated ✩

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

customer buys 17.01 in gas and requests one five dollar [$5] lottery ticket, two one dollar [$1] lottery tickets, and one [$3]

Anettt [7]

100 - 17.01 - 5 - 2 - 3 + 5 + 2 =
100 - 17.01 - 3 =
100 - 20.01 =
$ 79.99 <===

4 0

3 years ago

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