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2 years ago

12

For the problem I tried dividing but my answers were not correct. How can I solve this problem then? Can someone help me out her

e please?

1 answer:

astraxan [27]2 years ago

3 0

Answer:

8

Step-by-step explanation:

5 = 40

1 = x

Then we multiply by the rule of crisscrossing

5 x X = 40 x 1

5x = 40 then divide both by 5

X = 8

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Which ordered pair is a solution of the equation y = x - 2?

Liula [17]

Answer:

One order pair is

( 2,0)

Step-by-step explanation:

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Pick a value for

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8 0

3 years ago

Why are there so many bots on here?

Temka [501]

Answer:

Not sure, It's getting kinda annoying ngl

Step-by-step explanation:

4 0

2 years ago

PLEASE HELP! THIS IS FOR A UNIT!

AlexFokin [52]

Answer:

$n_{1} =-3+2i\\\\n_{2} =-3-2i$

Step-by-step explanation:

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7 0

3 years ago

Select the correct answer from each drop-down menu.

vazorg [7]

Answer:

Step 1: Distribute $-2$ to $5x$ and $8$

Step 2: Subtract from both sides of the equation $6x$

Step 3: Add to both sides of the equation $16$

Step 4: Divide both sides of the equation by $-16$

Step-by-step explanation:

Step 1: Apply the Distributive Property. Then you must distribute $-2$ to $5x$ and $8$

Then:

$-2(5x+8)=14+6x\\\\-10x-16=14+6x$

Step 2: You must apply the Subtraction property of Equality and subtract $6x$ from both sides of the equation. Then:

$-10x-16-6x=14+6x-6x\\\\-16x-16=14$

Step 3: You must apply the Addition property of Equality and add $16$ to both sides of the equation. Then:

$-16x-16+16=14+16\\\\-16x=30$

Step 4: You must apply the Division property of Equality and divide both sides by $-16$ . Then:

$\frac{-16x}{-16}=\frac{30}{-16}\\\\x=\frac{30}{-16}\\\\x=-\frac{15}{8}$

6 0

3 years ago

A standardized exam's scores are normally distributed. In a recent year, the mean test score was 14901490 and the standard dev

kicyunya [14]

<h2>Answer with explanation:</h2>

Given : A standardized exam's scores are normally distributed.

Mean test score : $\mu=1490$

Standard deviation : $\sigma=320$

Let x be the random variable that represents the scores of students .

z-score : $z=\dfrac{x-\mu}{\sigma}$

We know that generally , z-scores lower than -1.96 or higher than 1.96 are considered unusual .

For x= 1900

$z=\dfrac{1900-1490}{320}\approx1.28$

Since it lies between -1.96 and 1.96 , thus it is not unusual.

For x= 1240

$z=\dfrac{1240-1490}{320}\approx-0.78$

Since it lies between -1.96 and 1.96 , thus it is not unusual.

For x= 2190

$z=\dfrac{2190-1490}{320}\approx2.19$

Since it is greater than 1.96 , thus it is unusual.

For x= 1240

$z=\dfrac{1370-1490}{320}\approx-0.38$

Since it lies between -1.96 and 1.96 , thus it is not unusual.

5 0

3 years ago