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3 years ago

Write a ratio, in simplest form and a percent for the shaded area

Mathematics

2 answers:

musickatia [10]3 years ago

4 0

Answer:

do you have an image

Step-by-step explanation:

mars1129 [50]3 years ago

3 0

Answer:

drhjrdrhj

Step-by-step explanation:

You might be interested in

James Wrote the number 8,980,000 in scientific notation which number did he write

Colt1911 [192]

<span>8,980,000 = 8.98 x 10</span>⁶

6 0

3 years ago

Given the equation 4x2 − 8x + 20 = 0, what are the values of h and k when the equation is written in vertex form a(x − h)2 + k =

GenaCL600 [577]

Answer:

The correct option is:

h = 1, k = 16

Step-by-step explanation:

y=4x^2-8x+20 =0

It is a quadratic formula in standard form:

ax^2+bx+c

where a = 4 , b = -8 and c=20

The vertex form is:

a(x − h)2 + k = 0

h is the axis of symmetry and (h,k) is the vertex.

Calculate h according to the following formula:

h = -b/2a

h= -(-8)/2(4)

h = 8/8

h = 1

Substitute k for y and insert the value of h for x in the standard form:

ax^2+bx+c

k = 4(1)^2+(-8)(1)+20

k = 4-8+20

k=-4+20

k = 16

Thus the correct option is h=1, k=16....

8 0

3 years ago

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points. Rachel scored 78 p

Artemon [7]

Answer:

Keenan's z-score was of 0.61.

Rachel's z-score was of 0.81.

Step-by-step explanation:

Z-score:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Keenan scored 80 points on an exam that had a mean score of 77 points and a standard deviation of 4.9 points.

This means that $X = 80, \mu = 77, \sigma = 4.9$