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2 years ago

Help me asap I'll mark brainliest!!!!!!!!!

Mathematics

1 answer:

pshichka [43]2 years ago

3 0

Answer:

NO!!!! NO!!!! NOOOOOO!!!!!!!

Step-by-step explanation:

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Find the area of the circle from the given radius.

steposvetlana [31]

The answer should be b. 12

3 0

3 years ago

Thanks to who ever helped me with this question!

ollegr [7]

Answer:

-9

Step-by-step explanation:

Since the opposite angles of both sides are congruent we know that the sides must be congruent also. Therefore, to solve for x we can set them equal to each other.

So the equation will be x+19=10, to solve we must isolate the variable.

First, subtract 19 from both sides

x+19-19=10-19

Then, simplify

x=-9

3 0

3 years ago

Read 2 more answers

Need help ASAP!!!!!!!!!!

Marysya12 [62]

Answer:

the last two

Step-by-step explanation:

8 0

3 years ago

Refer to the previous exercise. The researchers wanted a sufficiently large sample to be able to estimate the probability of pre

Maru [420]

Answer:

The large sample n = 117.07

Step-by-step explanation:

Step(i):-

Given that the estimate error (M.E) = 0.08

The proportion (p) = 0.75

q =1-p = 1- 0.75 =0.25

Level of significance = 0.05

Z₀.₀₅ = 1.96≅ 2

Step(ii):-

The Marginal error is determined by

M.E = $\frac{Z_{0.05} \sqrt{p(1-p)} }{\sqrt{n} }$

$0.08 = \frac{2 X \sqrt{0.75(1-0.75)} }{\sqrt{n} }$

Cross multiplication , we get

$\sqrt{n} = \frac{2 X \sqrt{0.75(1-0.75)} }{0.08 }$

√n = $\frac{2 X0.4330}{0.08} = 10.825$

squaring on both sides , we get

n = 117.07

Final answer:-

The large sample n = 117.07

6 0

2 years ago

US average math SAT scores follow a normal distribution with a mean of 505 and a standard deviation of 112. A sample of 64 enter

Hitman42 [59]

Answer:

The claim that the scores of UT students are less than the US average is wrong

Step-by-step explanation:

Given : Sample size = 64

Standard deviation = 112

Mean = 505

Average score = 477

To Find : Test the claim that the scores of UT students are less than the US average at the 0.05 level of significance.

Solution:

Sample size = 64

n > 30

So we will use z test

$H_0:\mu \geq 477\\H_a:\mu < 477$

Formula : $z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

$z=\frac{505-477}{\frac{112}{\sqrt{64}}}$

$z=2$

Refer the z table for p value

p value = 0.9772

α=0.05

p value > α

So, we accept the null hypothesis

Hence The claim that the scores of UT students are less than the US average is wrong

3 0

3 years ago