Answer:
We accept the null hypothesis and conclude that voltage for these networks is 232 V.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 232 V
Sample mean,
= 231.5 V
Sample size, n = 66
Sample standard deviation, s = 2.19 V
Alpha, α = 0.05
First, we design the null and the alternate hypothesis
We use Two-tailed t test to perform this hypothesis.
Formula:
Putting all the values, we have
Now,
Since,

We accept the null hypothesis and conclude that voltage for these networks is 232 V.
Find the cost for one ticket for the science center.
36.75 ÷ 3 = 12.25
Multiply the product with 12 to get the cost of 12 tickets.
12.25 × 12 = 147
Find the cost of 1 zoo ticket.
51 ÷ 4 = 12.75
Find the cost of 12 tickets by multiplying the product and 12
12.75 × 12 = $153
12 Science Center Tickets cost $147
12 zoo tickets cost $153
The <u><em>Zoo</em></u> costs more than the Science Center.
To find out how much more the Zoo costed, subtract 147 from 153.
153 - 147 = 6
The zoo only costs <u><em>$6 </em></u>more for 12 students.
So, you're answers are Zoo and $6
Answer:
<em>x</em> = 75
Step-by-step explanation:
Since if you could take the second line on the line and move it down a bit it would be the same angle and so all you have to do is 90-15 = 75 so x=75.
P.S (I'm sorry if this didn't help :( )
The measure of m<MAR= 28 de
<h3>What is angle bisector?</h3>
Angle bisector in geometry refers to a line that splits an angle into two equal angles. Bisector means the thing that bisects a shape or an object into two equal parts.
Given:
AT bisects <MAR
and, m<MAT= (6x-4) and and m<RAT = (2x+8)
As, <MAR is bisected by ray AT.
So,
<MAT = <RAT
6x-4= 2x+8
6x- 2x = 8+4
4x = 12
x= 3
So, <MAT = 6x-4= 6*3-4= 14
and, <RAT = 2x+8 = 14
Hence, MAR= <MAT + <RAT = 14+14=28
Learn more about angle bisector here:
brainly.com/question/28557178
#SPJ4
Answer:
choice c
Step-by-step explanation:
if you divide, subtract, and or add it would be a decimal with a fraction and the answer would be 20.9458383 so c is most accurate