Answer:
8
x
10
^0
Step-by-step explanation:
the reason why is because there is no 0 behind the 8 so the exponent could not say a number so it is just 0
The graph third represents the provided situation first line increases and then constant then again increases and becomes constant.
<h3>What is a line graph?</h3>
A line graph is a graph made up of segments of straight lines that connect the depicted data points.
We have:
Keisha ran to swim practice but needed to stop by her friend Natasha’s house on the way. She stayed at Natasha’s house for 15 minutes, and together they ran the rest of the way to practice, where they stayed for 1 hour.
Based on the information given we can say graph third represents the situation.
First Keisha ran to swim practice so the line first increases and then she but needed to stop by her friend Natasha’s house on the way. She stayed at Natasha’s house for 15 minutes which means the line should be constant.
Again they both ran so line increases.
Then they both stayed for 1 hour, so the line should be constant.
Thus, the graph third represents the provided situation first line increases and then constant then again increases and becomes constant.
Learn more about the line chart here:
brainly.com/question/23680294
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12⁷/₁₂ - 5¹⁴/₅
12⁷/₁₂ - 7⁴/₅
12³⁵/₆₀ - 7⁴⁸/₆₀
11⁹⁵/₆₀ - 7⁴⁸/₆₀
4⁴⁷/₆₀
Answer:
he probability that the sample mean deviates from the population mean = 112 by no more than 4= 0.9050
Step-by-step explanation:
mean = 112
sd = 12/sqrt(25) = 2.4
z = (x - μ)/σ
z = (116 - 112)/2.4 = 1.67
Therefore,
P(X >= 116) = P(z <= (116 - 112)/2.4)
= P(z >= 1.67)
= 1 - 0.9525
= 0.0475
z = (x - μ)/σ
z1 = (108 - 112)/2.4 = -1.67
z2 = (116 - 112)/2.4 = 1.67
Therefore, we get
P(108 <= X <= 116) = P((116 - 112)/2.4) <= z <= (116 - 112)/2.4)
= P(-1.67 <= z <= 1.67) = P(z <= 1.67) - P(z <= -1.67)
= 0.9525 - 0.0475
= 0.9050
X + 2y = 1 ______×1
4x + y = 11 _____ ×2
x + 2y = 1
8x + 2y = 22
8x + 2y = 22
(-)
x + 2y = 1
_____________
7x = 21
x = 3
3 + 2y = 1
2y = 2
y = 1
i am a mathematics teacher. if anything to ask please pm me
