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hodyreva [135]
3 years ago
5

Find the area of the shaded region under the standard distribution curve.

Mathematics
2 answers:
docker41 [41]3 years ago
8 0

Answer:

D. 0.7611

Step-by-step explanation:

We have been given a graph of a normal standard distribution curve. We are asked to find the area of the shaded region under the given standard distribution curve.

The area of the shaded region under the standard distribution curve would be area of a z-score of 1.60 minus area of a z-score of -0.90 that is P(-0.90

Using normal distribution table, we will get:

P(-0.90

P(-0.90

Therefore, the shaded area under the curve is 0.7611 and option D is the correct choice.

Yanka [14]3 years ago
3 0

Answer:

D. 0.7611

Step-by-step explanation:

The area is:

P(z<1.60) − P(z<-0.90)

Looking up the values in a z-score table:

0.9452 − 0.1841

0.7611

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Read 2 more answers
Element X is a radioactive isotope such that its mass decreases by 26% every day. If an experiment starts out with 810 grams of
ioda

Answer:

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance is given by:

A(t) = A(0)(1-r)^t

In which A(0) is the initial amount and r is the decay rate, as a decimal.

Hourly rate of change:

Decreases 26% by day. A day has 24 hours. This means that A(24) = (1-0.26)A(0) = 0.74A(0); We use this to find r.

A(t) = A(0)(1-r)^t

0.74A(0) = A(0)(1-r)^{24}

(1-r)^{24} = 0.74

\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}

1 - r = (0.74)^{\frac{1}{24}}

1 - r = 0.9875

r = 1 - 0.9875 = 0.0125

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

Starts out with 810 grams of Element X

This means that A(0) = 810

Element X is a radioactive isotope such that its mass decreases by 26% every day.

This means that we use, for this equation, r = 0.26.

The equation is:

A(t) = A(0)(1-r)^t

A(t) = 810(1 - 0.26)^t

A(t) = 810(0.74)^t

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

5 0
3 years ago
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