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3 years ago

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Suppose the scores on a history quiz are normally distributed with a mean of 73 and a standard deviation of 4. Which group descr

ibes 16% of the population of history quiz scores?
A. scores below 65

B. scores below 75

C. scores above 77

D. scores above 81

1 answer:

KiRa [710]3 years ago

8 0

B is the correct answer the score is below 75

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Line CD passes through points C(3,-5) and D(6,0). What is the equation of line CD in standard form?

Serga [27]

Answer:

hope this helps

the explanation is in the picture

please like and Mark as brainliest

3 0

3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago

Identify the domain of the function shown in the graph.

kompoz [17]

<u>Answer:</u>

The answer is A choice.

<u>Explanation:</u>

By looking at the graph, we notice that at the left side, the domain/x-value stops at -5 and 5.

Thus, the domain is at <u>the left end side to the right end side.</u>

The domain stops at -5 on the left side and 5 on the right side. Thus the answer is

-5 < x < 5.

The A choice.

5 0

2 years ago

Read 2 more answers

Use the diagram to complete these statements. If m26 = 50°, then m2 = 50° If m 25 = 70°, then m = 70 What is the measure of 22?

Thepotemich [5.8K]

Answer: 1,3,60°

Step-by-step explanation:

4 0

3 years ago

Toni had 5 1/2 kilograms of sugar.how many cakes can he bake if he ise 1/4 hilogram of sugar per cake?

allochka39001 [22]

Answer:

22 cakes

Step-by-step explanation:

5 1/2kg = 11/2 = 22/4

22/4 is what you have in sugar while you only need 1/4 per cake, so you can make 22 cakes

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3 years ago

Read 2 more answers