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DanielleElmas [232]
3 years ago
11

A placement test is administered to collegiate freshmen. The test is known to be approximately normal with a mean of 65 and a st

andard deviation of 5. Sheila scored a 59 on this test. Does this score place Sheila in the bottom 10% of test takers?

Mathematics
1 answer:
rosijanka [135]3 years ago
5 0

Answer: D. She is above 11.5% of test takers

Step-by-step explanation:

So logically we can eliminate like 3 answers and let me explain why

So first we can eliminate both A and B because the question asks "Does this score place Sheila in the bottom 10% of test takers?" If the answer were<u> yes </u>she wouldn't be "above 8.5%" or "above 11.5%" because she would be below.

So that would leave both C and D right. Lets look at C first. It says "No, she is above 8.5% of test takers" But if she is above 8.5% she could hypothetically be in the 9% or the 9.5% of test takers. Since this answer doesn't gurantee she would be above 10% we can't choose it. Now D it says "No, she is above 11.5% of test takers" This is correct because She is not in the bottom 10% because according to the empirical rule she is I would say around or just under 16% of test takers. Also the 11.5% part gurantees that anything above the number 11.5 is greater than 10%. Unlike in C which only says the number is greater than 8.5 which means the number could be 8.6 which is NOT greater than 10.

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Help me plssssss i don’t know how to do this
makkiz [27]

Answer:

AC = \sqrt{15}

Step-by-step explanation:

Assuming you require AC

Using Pythagoras' identity in the right triangle

The square on the hypotenuse is equal to the sum of the squares on the other 2 sides , that is

AC² + BC²= AB²

AC² + 7² = 8²

AC² + 49 = 64 ( subtract 49 from both sides )

AC² = 15 ( take square root of both sides )

AC = \sqrt{15} ≈ 3.87 ( to 2 dec. places )

7 0
3 years ago
Find x Round to the nearest tenth
nikdorinn [45]

Answer:

15.5

Step-by-step explanation:

So the formula here is the square of the tangent length is equal to the product of the exterior part of the secant line and total length of the secant.

So that means we have x^2=10(14+10).

After simplifying the right hand side we have the equation is x^2=10(24) or x^2=240.

To get rid of the square on the x, we square root both sides.

x=sqrt(240)

x=15.49193338

To the nearest tenths, the answer is x=15.5

So 15.5 inches is the length of x.

3 0
3 years ago
Grade 4c left their school at7:30 am for their trip to the market. They return at 10:45 am how much time did they spend on the t
SOVA2 [1]
3:15

Step by step explanation:

10:45
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———
3:15
6 0
2 years ago
In △ABC, point M is the midpoint of AB , point D∈ AC so that AD:DC=2:5. If AABC=56 yd2, find ABMC, AAMD, and ACMD.
Komok [63]

Since point M is the midpoint of AB, then AM=MB.

Consider the area of the triangles ABC and BMC:

A_{ABC}=\dfrac{1}{2}\cdot AB\cdot h_c=56\ yd^2,

where h_c is the height drawn from the vertex C to the side AB.

So, AB\cdot h_c=112\ yd^2.

Now

A_{BMC}=\dfrac{1}{2}\cdot BM\cdot h_c=\dfrac{1}{2}\cdot \dfrac{AB}{2}\cdot h_c=\dfrac{1}{4}\cdot AB\cdot h_c=\dfrac{1}{4}\cdot 112=28\ yd^2.

Also

A_{AMC}=A_{ABC}-A_{BMC}=56-28=28\ yd^2.

Now consider the area of the triangles AMD and CMD. Let h_M be the height drawn from the point M to the side AC.

A_{AMD}=\dfrac{1}{2}\cdot AD\cdot h_M=\dfrac{1}{2}\cdot \dfrac{2AC}{7}\cdot h_M=\dfrac{2}{7}\cdot \left(\dfrac{1}{2}\cdot AC\cdot h_M\right)=\dfrac{2}{7}\cdot A_{AMC}=\dfrac{2}{7}\cdot 28=8\ yd^2.

Therefore,

A_{MDC}=A_{AMC}-A_{AMD}=28-8=20\ yd^2.

Answer: A_{MBC}=28\ yd^2, A_{AMD}=8\ yd^2, A_{MDC}=20\ yd^2.

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