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UkoKoshka [18]
3 years ago
10

GEOMETRY PLS HELP!!!

Mathematics
1 answer:
Ymorist [56]3 years ago
7 0

Answer:

PA = 17

Step-by-step explanation:

The centroid is the point of intersection of the 3 medians of the triangle.

A median is is a segment that goes from one of the triangles vertices to the midpoint of the opposite side.

QA is a median , so bisects PR at A , then

PA = 0.5PR = 0.5 × 34 = 17

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6/1000 as a common fraction
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Evaluate the expression when d = 20 and v = 19 20v + 12d - 6v
Alexeev081 [22]
D = 20 , v = 19
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= 20 (19) + 12 (20) - 6 (19)
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What is 12% on $3,200 in sales?
Karolina [17]

Answer:

The easiest way of calculating discount is, in this case, to multiply the normal price $3200 by 12 then divide it by one hundred. So, the discount is equal to $384. To calculate the sales price, simply deduct the discount of $384 from the original price $3200 then get $2816 as the sales price.

4 0
3 years ago
A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le
CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁:  μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}  } }

Z≈N(0;1)

Z= \frac{(533-340)-120}{\sqrt{\frac{84^2}{56} +\frac{93^2}{102}  } }= 5.028

I hope this helps!

3 0
4 years ago
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