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3 years ago

7th grade math guys!! Proportion equation?? anyone help please???!!

Mathematics

2 answers:

Aleks [24]3 years ago

7 0

The answer is B.

I need 20 character for this answer

Virty [35]3 years ago

6 0

Hey there!

In this problem, we're looking for x's <em>relation</em> to y. We want to know what we multiply, divide, or do any other operation to on x to get y.

And that, is why we have algebra.

We take our x value, and we put a variable, z, after it. That allows us to solve for z and see what we are doing to x to get y.

Let's try one:

We have 2, and when it goes through our function, we get 16. We have:

2z = 16
z = 8

Next one:

3z = 24
z = 8

Finally:

4z = 32
z = 8

We see a common pattern. Y is just x multiplied by 8.

Therefore, your answer is B, y = 8x

Hope this helps!

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Jenny has some tiles in a bag. The tiles are of three different colors: purple, pink, and orange. Jenny randomly pulls a tile ou

nlexa [21]

Answer: 26/50

Step-by-step explanation:

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3 years ago

How do I solve this???

Bezzdna [24]

I hope this helps you

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4 years ago

Four cards are drawn from a standard deck. No cards are returned.

AnnyKZ [126]

<span>Event A is more likely...Pr(A)=(16/52)(15/51)(14/50)(13/49) .... and Pr(B)= (4/52)(3/51)(2/50)(1/49)</span>

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3 years ago

The original price for the camera is 150 if a discount of 30% is given, what is the discounted price

bazaltina [42]

Hello,

The camera is $150

The discount is 30%

First multiply the % by 100 to get a decimal

150 * 0.30 = 45

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The Camera now costs $105

The discount price was $45

Hope this helps

-GoldenWolfX

3 0

3 years ago

Read 2 more answers

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that $n = 723, p = 0.037$ .

So for the approximation, we will have:

$\mu = E(X) = np = 723*0.037 = 26.751$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08$

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 15 - 0.5) = P(X \geq 14.5)$ , which is 1 subtracted by the pvalue of Z when X = 14.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14.5 - 26.751}{5.08}$

$Z = -2.41$

$Z = -2.41$ has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, $P(X \geq 30 - 0.5) = P(X \geq 29.5)$ , which is 1 subtracted by the pvalue of Z when X = 29.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{29.5 - 26.751}{5.08}$

$Z = 0.54$

$Z = 0.54$ has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946 = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, $P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5)$ , which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So