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vlabodo [156]
3 years ago
14

How is an easy way to solve proportions with variables on both sides

Mathematics
1 answer:
tekilochka [14]3 years ago
3 0

m+11/11=4m-2/33

m+11(33)=4m-2(11)

33m+363=44m-22

33m-44m=-22-363

-11m=-385

m=35

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If we want to find the distance between two points on a number line use the distance formula:


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What is the answer of K < -5 ?
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To your question, "K<-5", that is the answer. So the answer is K<-5
8 0
3 years ago
Please help me plzzzzzz​
erma4kov [3.2K]

Answer:

976 square cm

Step-by-step explanation:

Area of one 2d triangle face: (24*14)/2=168 square cm

Area of both 2d triangles faces: 168 + 168 =336 square cm

Area of one side: 25*10=250 square cm

Area of both sides: 250+250=500

Area of base: 14*10=140 (since it's a triangular prism, there's only one base)

Surface Area: 336+500+140=976 square cm

Hope this helps! :)

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3 years ago
A shopping mall charges $30 per vehicle plus $2 per person as an admission fee. The total charge for a vehicle with x people is
Alecsey [184]
Changing the 30 to 10 is a 20 unit drop ( 30-10=20)
 so the line would shift vertically down by 20

<span>The line will shift vertically down by $20.</span>
7 0
3 years ago
The strength of a certain type of rubber is tested by subjecting pieces of the rubber to an abrasion test. For the rubber to be
My name is Ann [436]

Answer:

Step-by-step explanation:

Hello!

You have the hypothesis that the average weight loss for rubber after an abrasion test is less than 3.5 mg. To test this a large sample of pieces of rubber were sampled and subjected to the abrasion test.

With the given information you must test whether the researcher's hypothesis is sustained or not.

The study variable is,

X: Weight loss of rubber cured in a certain way after being subjected to the abrasion test. (mg)

There is no information about the variable distribution, but since it is said that the sample is a "large number" I'll take it as if it is bigger than 30 and apply the Central Limit Theorem to use the approximation of the sample mean to normal. This way I can use the Z-statistic for the test.

Symbolically the statistic hypothesis is:

H₀: μ ≥ 3.5

H₁: μ < 3.5

α: 0.05 (since is not listed, I'll choose one of the most common signification levels)

You have a one-tailed critical region, this means the p-value will also be one-tailed to the left of the distribution (i.e. →-∞)

The formula of the statistic is:

Z= <u> X[bar] - μ </u> ≈ N(0;1)

        δ/√n

To calculate the statistic you have to use the information given.

The sample mean X[bar]= 3.4 mg

Upper bond of 95% CI= 3.45 mg

The basic structure of a CI for the mean is

"estimator" ± "margin of error"

Upper bound is "estimator" + "margin of error"

Using the formula:

Ub= X[bar] + d ⇒ 3.45= 3.4 + d

⇒ d= 3.45 - 3.4 = 0.05

Where d is the margin of error

d= Z_{1-\alpha /2} * (δ/√n)

d= Z_{0.975} * (δ/√n)

d/Z_{0.975}= (δ/√n)

(δ/√n)= 0.05/ 1.96 = 0.0255

(δ/√n) is the denominator in the formula, corresponds to the standard deviation of the distribution.

Now you have all values and can calculate the statistic under the null hypothesis:

Z= <u> 3.4 - 3.5 </u> = -3.92

       0.0255

And the p-value:

P(Z ≤ -3.92) = 0.000044 ⇒ My Z- table goes up to P(Z ≤ -3.00) = 0.001, so using strictly the table I can say that the probability is less than 0.001.

To calculate the exact probability I've used a statistic program.

p-value < 0.001

I hope it helps!

8 0
3 years ago
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