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3 years ago

Please help. will mark brainliest if its right.180-x=3(90-x)

Mathematics

1 answer:

DerKrebs [107]3 years ago

4 0

<h2>hope it helps you please mark me as brainliest </h2>

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I NEED THE ANSWER PRONTO CUZ I'M DOING THE TEST RIGHT NOW!!!!!Devon made the scatter plot to track his bank balance over time.

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It should be <span>y=15x+250</span>

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3 years ago

Solve 2x + 4 > 16<br><br> A. x > 10<br> B. x < 6<br> C. x < 10<br> D. x > 6

Salsk061 [2.6K]

Hey there! :)

2x + 4 > 16

We're simply just trying to get x onto it's own side, so let's get rid of any numbers on the same side of it.

We can start out by subtracting both sides by 4.

2x + 4 - 4 > 16 - 4

Simplify.

2x > 12

Then, divide both sides by 2.

2x ÷ 2 > 12 ÷ 2

Simplify.

x > 6

Therefore, our answer is D. x > 6

~Hope I helped!~

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2 years ago

Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean

Zinaida [17]

Answer:

$E[R]$ = 99 Ω

$\sigma_R$ = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, $E[R] = \frac{a+b}{2} = \frac{95+103}{2} = \frac{198}{2}$ = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

$V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333$

Hence,

The standard deviation of resistance, $\sigma_R = \sqrt{V(R)} = \sqrt5.333$ = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

$z_1 = \frac{102-99}{2.3094} = 1.299$

$z_2 = \frac{98-99}{2.3094} = -0.433$

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

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3 years ago

Compatible numbers to estimate 75 plus 26

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101 is the correct answer for 75+26

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2 years ago

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Is 15/20 a half or a whole?

Arte-miy333 [17]

Answer:

15/20 is neither its 3-quarters

Step-by-step explanation:

15/20 is 3/4 which is 75% therefore...

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2 years ago

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