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3 years ago

I WILL GIVE BRAINLIEST!

Mathematics

2 answers:

Kitty [74]3 years ago

3 0

Answer:

A. reflection over the y-axis followed by reflection over the x-axis

Step-by-step explanation:

Thank me later :))

Charra [1.4K]3 years ago

3 0

Step-by-step explanation:

A. Rreflection over the y-axis followed by reflection over the x-axis

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Divide: 3/7 ÷ 1/2 A) 116 B) 67 C) 25 D) 314

labwork [276]

The answer is B) 6/7

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2 years ago

Read 2 more answers

If x2 + y2 = 38 and x - y = 2, then what is x? If there are two possible answers, then enter the larger of the two.

Ket [755]

If x-y=2 then x=2+y

And if you insert that in the first equation

$(2 + y) {}^{2} + y {}^{2} = 38$
$4 + 4y + y {}^{2} + {y}^{2} = 38$
$2y {}^{2} + 4y - 34 = 0$
$y = - 1 + \sqrt{17}$
Or
$y = - 1 - \sqrt{17}$

The first one is bigger so the answer is

$y = - 1 - \sqrt{17}$

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3 years ago

Read 2 more answers

RAH SOMEONE PLZ HELP ASAP/

lesantik [10]

1rst step is 7×22=154

2nd step is 8-2=6

3rd step is 154+6=160

Last step is 160÷3=53.3333333333

I hope this helps it will be b because 53.3333333333 is near 60

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3 years ago

PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!

KonstantinChe [14]

Step-by-step explanation:

RS = 8 × cos(61°) = 2.9 [1 dp] = 3cm

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3 years ago

SAT scores are normally distributed with a mean of 1,500 and a standard deviation of 300. An administrator at a college is inter

Rama09 [41]

Answer:

The administrator should sample 968 students.

Step-by-step explanation:

We have to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.88}{2} = 0.06$

Now, we have to find z in the Z-table as such z has a p-value of $1 - \alpha$ .

That is z with a p-value of $1 - 0.06 = 0.94$ , so Z = 1.555.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

Standard deviation of 300.

This means that $n = 300$

If the administrator would like to limit the margin of error of the 88% confidence interval to 15 points, how many students should the administrator sample?

This is n for which M = 15. So

$M = z\frac{\sigma}{\sqrt{n}}$

$15 = 1.555\frac{300}{\sqrt{n}}$