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3 years ago

I need help with this question

Mathematics

1 answer:

Yanka [14]3 years ago

4 0

Answer:

f(g(4)) = 5

Step-by-step explanation:

Recall that to find f(g(4)), we need to first find g(4) which we find via the given table:

g(4) = 0

and then use the graph to find f(0) = 5

Then f(g(4)) = 5

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Compare the fractions using the LCM. 3/5 and 1/4

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3 years ago

Find m<LMN if m<LMT=23 degree and m<TMN=144 degree

rodikova [14]

Answer:

∠LMN = 167°

Step-by-step explanation:

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3 years ago

A population of monkeys' tail lengths is normally distributed with a mean of 25 cm with a standard deviation of 8 cm. I am prepa

Crank

Answer:

The probability that the mean of my sample will be between 24 and 25 cm

P(24 ≤X⁻≤25) = 0.4772

Step-by-step explanation:

Step(i):-

Given mean of the Population 'μ'= 25c.m

Given standard deviation of the Population 'σ' = 8c.m

Given sample size 'n' = 256

Let X₁ = 24

$Z_{1} = \frac{x_{1}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{24-25}{\frac{8}{\sqrt{256} } } = -2$

Let X₂ = 25

$Z_{2} = \frac{x_{2}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{25-25}{\frac{8}{\sqrt{256} } } = 0$

Step(ii):-

The probability that the mean of my sample will be between 24 and 25 cm

P(24 ≤X⁻≤25) = P(-2≤ Z ≤0)

= P( Z≤0) - P(Z≤-2)

= 0.5 + A(0) - (0.5- A(-2))

= A(0) + A(2) ( ∵A(-2) =A(2)

= 0.000+ 0.4772

= 0.4772

Final answer:-

The probability that the mean of my sample will be between 24 and 25 cm

P(24 ≤X⁻≤25) = 0.4772

4 0

3 years ago

I need help with this question​

I need help with this question