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guapka [62]
3 years ago
12

I NEED HELP ASAP DUE IN 3 MINUTES WHOEVER ANSWERS IT RIGHT GETS BRIANIST

Mathematics
1 answer:
Damm [24]3 years ago
8 0

Answer:

It is D.

Step-by-step explanation:

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Find the value of Log2 32=y
lukranit [14]

Answer:

y=5

Step-by-step explanation:

We are solving

log_232=y

The logarithm is basically asking us this question:

2^y=32

Or what power should I raise 2 to so that it will give me 32?

We know that 2^5=32

Therefore y=5.

<em>P.S: calculating logarithms is not an algorithmic process: you have to rely mostly on educated guesses to get your  answers.</em>

3 0
3 years ago
Experiment: pick one number from 1 to 10. What is the sample space fo<br> this experiment?
aalyn [17]

Answer:

Step-by-step explanation:

Sample space = {1, 2 , 3 , 4 , 5, 6, 7, 8, 9, 10}

8 0
3 years ago
Add any three numbers of the following and answer should be 60?
mina [271]
Six, twenty- two, and thirty
3 0
3 years ago
Read 2 more answers
- 3y - 7=5y + 17 solve for y
KengaRu [80]

Answer: y=-3

Step-by-step explanation:

- 3y - 7=5y + 17

-7=8y+17

8y=-24

y=-3

8 0
2 years ago
A book claims that more hockey players are born in January through March than in October through December. The following data sh
astra-53 [7]

Answer:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

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

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

Step-by-step explanation:

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

H0: There is no difference of birthdates distributed throughout the​ year

H1: There is a difference between birthdates distributed throughout the​ year

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{total}{4}

And replacing we got:

E_{1} =\frac{67+56+30+37}{4}=47.5

And now we can calculate the statistic:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

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

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

3 0
4 years ago
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