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3 years ago

7

The exam scores of all 500 students were recorded and it was determined that these scores were normally distributed. If Jane's s

core is 0.8 standard deviation above the mean, then how many, to the nearest unit, students scored above Jane?
A. 394
B. 250
C. 400
D. 106
only answer if you know

2 answers:

olganol [36]3 years ago

7 0

The answer to this is D i did the math i am positive it is right

Wewaii [24]3 years ago

6 0

The answer is D I think if not sorry

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12. Find the distance between points P(-1, 2) and Q(-3, 4).

marusya05 [52]

Answer:

D=2.82

Step-by-step explanation:

Here we are given with two coordinates and asked to determine the distance between them.

Here we are going to use the distance formula, which is given as under

$D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Where

$(x_1,y_1) = (-1,2)$

$(x_2,y_2) = (-3,4)$

Replacing these values in the distance formula

$D =\sqrt{(-1-(-3))^2+(2-4)^2}$

$D =\sqrt{(-1+3)^2+(-2)^2}$

$D =\sqrt{(2)^2+4}$

$D =\sqrt{4+4}$

$D =\sqrt{8}$

$D= 2.8284$

Hence the Distance is 2.82 units

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3 years ago

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3 years ago

The area of a rectangular field is 5695 m².

denpristay [2]

Answer:

85 m

Step-by-step explanation:

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2 years ago

Which is the inverse of the function a(d)=5d-3? And use the definition of inverse functions to prove a(d) and a-1(d) are inverse

Drupady [299]

Answer:

$a'(d) = \frac{d}{5} + \frac{3}{5}$

$a(a'(d)) = a'(a(d)) = d$

Step-by-step explanation:

Given

$a(d) = 5d - 3$

Solving (a): Write as inverse function

$a(d) = 5d - 3$

Represent a(d) as y

$y = 5d - 3$

Swap positions of d and y

$d = 5y - 3$

Make y the subject

$5y = d + 3$

$y = \frac{d}{5} + \frac{3}{5}$

Replace y with a'(d)

$a'(d) = \frac{d}{5} + \frac{3}{5}$

Prove that a(d) and a'(d) are inverse functions

$a'(d) = \frac{d}{5} + \frac{3}{5}$ and $a(d) = 5d - 3$

To do this, we prove that:

$a(a'(d)) = a'(a(d)) = d$

Solving for $a(a'(d))$

$a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})$

Substitute $\frac{d}{5} + \frac{3}{5}$ for d in $a(d) = 5d - 3$

$a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3$

$a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3$

$a(a'(d)) = d + 3 - 3$

$a(a'(d)) = d$

Solving for: $a'(a(d))$

$a'(a(d)) = a'(5d - 3)$

Substitute 5d - 3 for d in $a'(d) = \frac{d}{5} + \frac{3}{5}$

$a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}$

Add fractions

$a'(a(d)) = \frac{5d - 3+3}{5}$

$a'(a(d)) = \frac{5d}{5}$

$a'(a(d)) = d$

Hence:

$a(a'(d)) = a'(a(d)) = d$

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2 years ago

What is the graph of the solution to the following compound inequality?

podryga [215]

Answer:

1 lap in 8/10 (4/5) of a minute

Answer is B, 8/10 minute! or 4

⭐ Please consider brainliest! ⭐

✉️ If any further questions, inbox me! ✉️

Step-by-step explanation:

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3 years ago