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Westkost [7]
2 years ago
9

The product of the length of 2 unequal poles before cutting 2 cm from each was 285 cm. Their present sum after cutting 2cm from

each is 30cm. What is the length of the shorter pole? The product of the length of 2 unequal poles before cutting 2 cm from each was 285 cm . Their present sum after cutting 2cm from each is 30cm . What is the length of the shorter pole ?​
Mathematics
1 answer:
labwork [276]2 years ago
8 0

Answer:

15 cm

Step-by-step explanation:

The length of the shorter pole can be found by forming and subsequently solving 2 equations.

Start by defining the variables that are going to be used in the working.

Let the original length of the shorter pole be a cm and that of the longer pole be b cm.

'Product' refers to the multiplication operation.

ab= 285 -----(1)

On the other hand, 'sum' refers to the addition operation.

Length of shorter pole after cutting= a -2

Length of longer pole after cutting= b -2

(a -2) +(b-2)= 30

a +b -4= 30

Adding 4 to both sides:

a +b= 30 +4

a +b= 34 -----(2)

From (2):

a= 34 -b -----(3)

Let's solve by substitution:

Substitute (3) into (1):

b(34 -b)= 285

Expand:

34b -b²= 285

b² -34b +285= 0

Factorise:

(b -15)(b -19)= 0

b -15= 0       or       b -19= 0

b= 15            or             b= 19

Substitute into (1):

a(15)= 285       or       a(19)= 285

a= 285 ÷15      or             a= 285 ÷19

a= 19                or             a= 15

Since a <b, a= 15 and b= 19.

Thus, the length of the shorter pole is 15 cm.

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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
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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.

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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!

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