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julia-pushkina [17]
3 years ago
10

The base of a triangle is 2 more than twice the height of the triangle. Find the measure of the base and the height if the area

is 20 ft^2 (the height is 4 and the base is 10, show a process so I can see how you got there
Mathematics
1 answer:
vekshin13 years ago
8 0
B = 2 + 2H 
A = 20 ft^2
H = H

Since the Area of a Triangle is Base * Height then the formula should be:
A = BH
20 = H(2+2H)
20 = 2H^2 + 2H
0 = 2H^2 + 2H - 20 

Find the zeros of the quadratic equation through substituting the values of a,b, and c. You will get the zeros of x = 2.7 and -3.7. The only thing that would make sense is you use the x=2.7 because there are no negative heights.

Since b = 2h+2 
b= 2(2.7) + 2
b= 5.4 + 2 
b=7.4

I don't know where you got 4 and 10, but what I got is different. 
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A rectangle has length 4 yards more than its width. The area of the rectangle is 96 square yards. Find the
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Answer:

Length = 12

Width = 8

Step-by-step explanation:

Area of a Rectangle = l x w

12 x 8 = 96

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2 years ago
− 50 67 ​ +1.5−100%
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Answer:

<em>Hello, I think the answer is -0.84 Hope That Helps!</em>

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3 years ago
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The pressure applied to a leverage bar varies inversely as the distance from the object. If 150 pounds is required for a distanc
AlexFokin [52]

Answer:

500 pounds

Step-by-step explanation:

Let the pressure applied to the leverage bar be represented by p

Let the distance from the object be represented by d.

The pressure applied to a leverage bar varies inversely as the distance from the object.

Written mathematically, we have:

p \propto \dfrac{1}{d}

Introducing the constant of proportionality

p = \dfrac{k}{d}

If 150 pounds is required for a distance of 10 inches from the object

  • p=150 pounds
  • d=10 inches

150 = \dfrac{k}{10}\\\\k=1500

Therefore, the relationship between p and d is:

p = \dfrac{1500}{d}

When d=3 Inches

p = \dfrac{1500}{3}\\\implies p=500$ pounds

The pressure applied when the distance is 3 inches is 500 pounds.

4 0
3 years ago
Find the midpoint of (0, 4) and (2, -2)
lidiya [134]

Answer:

\text{Midpoint} = (1, \; 1)

Step-by-step explanation:

M = (x_M, \; y_M)

M = \left(\dfrac{x_1 + x_2}{2}, \; \dfrac{y_1 + y_2}{2}\right)

M = \left(\dfrac{0 + 2}{2}, \; \dfrac{4 + -2}{2}\right)

M = \left(\dfrac{2}{2}, \; \dfrac{2}{2}\right)

M = (1, \; 1)

5 0
2 years ago
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A researcher compares two compounds (1 and 2) used in the manufacture of car tires that are designed to reduce braking distances
wariber [46]

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to compare two compounds, designed to reduce braking distance, used in tire manufacturing to prove if the braking distance of SUV's equipped with tires made with compound 1 is shorter than the braking distance of SUV's equipped with tires made with compound 2.

So you have 2 independent populations, SUV's equipped with tires made using compound 1 and SUV's equipped with tires made using compound 2.

Two samples of 81 braking tests are made and the braking distance was measured each time, the study variables are determined as:

X₁: Braking distance of an SUV equipped with tires made with compound one.

Its sample mean is X[bar]₁= 69 feet

And the Standard deviation S₁= 10.4 feet

X₂: Braking distance of an SUV equipped with tires made with compound two.

Its sample mean is X[bar]₂= 71 feet

And the Standard deviation S₂= 7.6 feet

We don't have any information on the distribution of the study variables, nor the sample data to test it, but since both sample sizes are large enough n₁ and n₂ ≥ 30 we can apply the central limit theorem and approximate the distribution of both variables sample means to normal.

The researcher's hypothesis, as mentioned before, is that the braking distance using compound one is less than the distance obtained using compound 2, symbolically: μ₁ < μ₂

The statistical hypotheses are:

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

α: 0.05

The statistic to use to compare these two populations is a pooled Z test

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } }

Z ≈ N(0;1)

Z_{H_0}= \frac{69-71-0}{\sqrt{\frac{108.16}{81} +\frac{57.76}{81} } }= -1.397

The rejection region if this hypothesis test is one-tailed to the right, so you'll reject the null hypothesis to small values of the statistic. The critical value for this test is:

Z_{\alpha  } = Z_{0.05}= -1.648

Decision rule:

If Z_{H_0} > -1.648 , then you do not reject the null hypothesis.

If Z_{H_0} ≤ -1.648 , then you reject the null hypothesis.

Since the statistic value is greater than the critical value, the decision is to not reject the null hypothesis.

At a 5% significance level, you can conclude that the average braking distance of SUV's equipped with tires manufactured used compound 1 is greater than the average braking distance of SUV's equipped with tires manufactured used compound 2.

I hope you have a SUPER day!

4 0
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