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likoan [24]
3 years ago
15

Can anyone help me with these

Mathematics
1 answer:
Studentka2010 [4]3 years ago
3 0
The answer is C

Trust me
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H(t) = –16t2 + 14t find the maximum height
guajiro [1.7K]
The height is represented as y and the time is represented as x. 


To find the maximum height (since this is a parabola that faces downwards) we need to find the vertex.


Vertex (x) = \frac{-(b)}{2a}

b = 14
a = -16

Substitute
\frac{-14}{2(-16)} =  \frac{-14}{-32}  =  \frac{7}{8}

Now plug 7/8 to t 
-16(7/8)^2 + 14(7/8)
-16(49/64) + 14(7/8)
-12.25 + 12.25
Answer: 0 ft
5 0
4 years ago
Read 2 more answers
I am having trouble with number 2. Please show how the answer was gotten.
Andrews [41]

We know how we got 90, since a right angle is 90 degrees,

180 degrees are in a triangle.

180-90=90

Find the pair that equals 90 so you get 180 degrees!

If you have more questions don’t have hesitation to ask!

Or add all number to get 180 degrees!

4 0
3 years ago
Find the mean of the following set of numbers. ​ ​ 8,5,9,5,5,7,3
Svetlanka [38]

Answer:

6

Step-by-step explanation:

add them all up and divide by 7

7 0
3 years ago
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Suppose y varies directly with x, and y = 24 when x = 8. What is the value of y when x = 10?
Solnce55 [7]
First, you want to set up a proportion.

8/24 = 10/y

Cross multiply to get

240 = 8y, then divide 8 from both sides to get y = 30.
6 0
3 years ago
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A production supervisor at a major chemical company wishes to determine whether a new catalyst, catalyst XA-100, increases the m
zheka24 [161]

Answer:

a. n= 47

b. n= 128

Step-by-step explanation:

Hello!

The objective of this experiment is to test if the new catalyst, XA-100, increases the mean hourly yield of a chemical process, that is known to be μ=750 (pounds per hour) with the current process.

You need to calculate the sample size to estimate the population with determined error margins.

To do so, since you have no population information, only that it is approximately normal distributed, you'll use the Student t statistic to get the sample size.

The formula of the margin of error (d) is:

d= t_{n-1: 1-\alpha/2} * (\frac{S}{\sqrt{n} })

I've  cleared the sample size of the formula

n= (S*\frac{t_{n-1; 1-\alpha /2} }{d} )^{2}

You need a sample size for the t-Student value and a standard deviation, that's why the information of a pilot study with n=5 and S= 19.62 is given.

a)

95% CI

d= 8 pounds

t_{n-1; 1-\alpha/2 } = t_{5-1;1-0.025}  = t_{4;0.975} =2.776

n= (19.62*\frac{2.776 }{8} )^{2}

n= 46.35 ≅ 47

b)

99% CI

d= 5 pounds

t_{n-1; 1-\alpha/2 } = t_{5-1;1-0.005}  = t_{4;0.995} =4.604

n= (19.62*\frac{4.604}{8} )^{2}

n= 127.49 ≅ 128

I hope it helps!

4 0
4 years ago
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