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3 years ago

Witch one is the right answer

Mathematics

2 answers:

frez [133]3 years ago

6 0

The answer is D good luck

S_A_V [24]3 years ago

6 0

If it has all equal sides it's equilateral. So the answer is A.

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The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,00

timama [110]

Answer: the probability that a randomly selected tire will have a life of exactly 47,500 miles is 0.067

Step-by-step explanation:

Since the life expectancy of a particular brand of tire is normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = life expectancy of the brand of tire in miles.

µ = mean

σ = standard deviation

From the information given,

µ = 40000 miles

σ = 5000 miles

The probability that a randomly selected tire will have a life of exactly 47,500 miles

P(x = 47500)

For x = 47500,

z = (40000 - 47500)/5000 = - 1.5

Looking at the normal distribution table, the probability corresponding to the z score is 0.067

6 0

3 years ago

How do you solve this problem

torisob [31]

Divide the price by the number of payments.

$15,480/36 = $430

4 0

3 years ago

F(x)=x^2 what is g(x)?<br><br> pls help me

Lady bird [3.3K]

G(x) = ax^2
y = ax^2
5 = a(1)^2
a = 5
Therefore, g(x) = 5x^2

8 0

3 years ago

What is the percent change of 20 - 60?

Sergeeva-Olga [200]

40% is the answer ...

3 0

2 years ago

Read 2 more answers

A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv

lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

of novels that seniors read during their time in college. An exit

survey was conducted with a random sample of 9 seniors. The

sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

been met, which of the following is a 94.645% confidence

interval for the population mean number of novels read by

all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

= n - 1

= 9 - 1

= 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

=1 - 0.94645

=0.05355 ≈ 0.05

⇒α = 0.5

Now,

$\frac{\alpha}{2} = \frac{0.05}{2}$

= 0.025

Then,

$t_{\frac{\alpha}{2}, df }$ = 2.306

∴ we get

Population mean = x⁻ ± $t_{\frac{\alpha}{2}, df }$ × $\frac{s}{\sqrt{n} }$

= 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

⇒Population mean = 7 ± 2.306 × $\frac{2.29}{\sqrt{9} }$

3 0

3 years ago

Read 2 more answers