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3 years ago

When considering a normal distribution of scores, what percentage of scores is equal to or below the mean

Mathematics

1 answer:

Rama09 [41]3 years ago

7 0

Answer:

50%

Step-by-step explanation:

If the distrubution is normal

and mean is the average score

that means that half of them should be equal to or below the mean

and the other half is equal to or above the mean

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Which of the following expressions represent “three times the difference between t and y

loris [4]

Answer:

3(t - y)

Step-by-step explanation:

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3 years ago

Х Convert fractions to percent 37/50 of a number is what percentage of that number?

Cloud [144]

Answer:

37/50 as a percent is 74%

Step-by-step explanation:

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3 years ago

Read 2 more answers

6x - 2y = 10

pishuonlain [190]

Answer:

y = 3x -5

Step-by-step explanation:

6x - 2y = 10

-6x. -6x

Inverse Operations

-2y = -6x + 10

÷-2. ÷-2. ÷-2

y = 3x - 5

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3 years ago

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What is the solution to the system of equations?

Burka [1]

Answer:

(10, -20).

Step-by-step explanation:

y= -5x + 30

x = 10

Substitute x = 10 in the first equation:

y = -5*10 + 30

y = -50 + 30

y = -20.

So the solution is (10, -20).

4 0

3 years ago

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Elena bikes 20 minutes each day for exercise. Write an equation to describe the relationship between her distance in miles, D, a

kompoz [17]

Answer:

The equation to describe the relationship between Elena distance:

a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

Solution:

Given, Elena bikes 20 minutes each day for exercise.

We have to write an equation to describe the relationship between her distance in miles, D, and her biking speed, in miles per hour,

We know that, distance travelled = speed $\times$ time

a. At a constant speed of 13 miles per hour for the entire 20 minutes

Her speed is 13 miles per hour.

Then, distance D miles = 13 miles per hour $\times$ 20 minutes

$\mathrm{D}=13 \text { miles per hour } \times \frac{20}{60} \text { hours } \rightarrow \mathrm{d}=13 \times \frac{1}{3} \rightarrow \mathrm{d}=4.34 \text { miles approximately. }$

b. At a constant speed of 15 minutes per hour for the first 5 minutes, then at 12 miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with 15 mph + distance travelled with 12 mph

$\begin{array}{l}{\mathrm{D}=15 \mathrm{mph} \times 5 \text { minutes }+12 \mathrm{mph} \times 15 \text { minutes }} \\\\ {\mathrm{D}=15 \mathrm{mph} \times \frac{5}{60} \text { hours }+12 \mathrm{mph} \times \frac{15}{60} \text { minutes }} \\\\ {\mathrm{D}=15 \times \frac{1}{12}+12 \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{5}{4}+3 \rightarrow \mathrm{D}=3+1.25 \rightarrow \mathrm{D}=4.25 \mathrm{miles}}\end{array}$

c. At a constant speed of M miles per hour for the first 5 minutes, then at N miles per hour for the last 15 minutes

Now, total distance travelled = distance travelled with M mph + distance travelled with N mph

$D = M mph \times 5 minutes + N mph \times 15 minutes$

$\mathrm{D}=\mathrm{M} \mathrm{mph} \times \frac{5}{60} \text { hours }+\mathrm{N} \mathrm{mph} \times \frac{15}{60} \text { minutes }$

$\mathrm{D}=\mathrm{M} \times \frac{1}{12}+\mathrm{N} \times \frac{1}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{N}{4} \rightarrow \mathrm{D}=\frac{M}{12}+\frac{3 N}{12} \rightarrow 12 \mathrm{D}=\mathrm{M}+3 \mathrm{N} \text { miles }$

Hence, a) D = 4.34 miles, b) D = 4.25 miles and c) 12D = M + 3N miles.

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4 years ago