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Elden [556K]
2 years ago
7

ASAP PLEASE ......................

Mathematics
1 answer:
jonny [76]2 years ago
4 0
Awnser 3 is the right one
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75% of of the number 124 is 93
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89 8989898989898989898

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6, 27, 45, 279.

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A humanities professor assigns letter grades on a test according to the following scheme. A: Top 12% of scores B: Scores below t
RUDIKE [14]

Answer:

The minimum score required for an A grade is 88.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 77.8, \sigma = 8.5

Find the minimum score required for an A grade.

Top 12%, which is at least the 100-12 = 88th percentile, which is the value of X when Z has a pvalue of 0.88. So it is X when Z = 1.175.

Z = \frac{X - \mu}{\sigma}

1.175 = \frac{X - 77.8}{8.5}

X - 77.8 = 1.175*8.5

X = 87.8

Rounding to the nearest whole number

The minimum score required for an A grade is 88.

5 0
2 years ago
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The area of a circle is increasing at the rate of pi cm^(2)/min. At what rate is the radius increasing when the area is 4pi cm^(
Vikentia [17]
This problem deals the rate of change.
For the formula of the area of a circle, we differentiate both sides with respect to time t.
(A = πr^2) d/dt
dA/dt = 2πr (dr/dt)

Since we don't know yet the radius r, the area of a circle is given.
A = πr^2
r^2 = A/π = 4π/π
r^2 = 4
r = 2 cm

Therefore, the rate of the radius is
dA/dt = 2πr (dr/dt)
dr/dt = (dA/dt)/(2πr)
dr/dt = π/(2π*2)
dr/dt = 0.25 cm/min

Hope this helps.
3 0
3 years ago
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