Answer: C
Step-by-step explanation:
4x + 1 = 9
4x + 1 -1 = 9-1
4x=8
8/4=2
Answer:
a) sample size is 40
Step-by-step explanation:
(a)
minutes
Margin of error, E is 75 seconds E=75/60= 1.25 minutes.
The level of significance
for 95% level of significance
For 95% confidence interval
from standard deviation table
Sample size required, n

Rounding off, n=40
Sample size =40
Keywords:
95% confidence, sample size, Wall Street Journal
The missing value x=3 is obtained by the concept of slope in coordinate geometry.
<h3>What is slope?</h3>
Positive or negative rates of change are both possible. Since the slope of a line is determined by the ratio of the vertical and horizontal change between any two locations on a plane or line, the slope is determined by the rise to run ratio. Where, Rise is the difference in elevation between two points. The horizontal shift between two places is known as a run.
Formula of slope =
(equation 1)
According to the question the coordinates are: xi=2, x2=x, y1=6, y2=10
Given, rate of change=4 (here rate of change=slope)
Putting all coordinates in equation 1
Slope=
4=4/x-2
x-2=1
x=3
Therefore, the missing value x=3
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Factor the following:
x^2 - 16 x + 63
The factors of 63 that sum to -16 are -7 and -9. So, x^2 - 16 x + 63 = (x - 7) (x - 9):
Answer: (x - 7) (x - 9)
<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
Formula: Heat = mass * specific heat * ΔT
Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C
=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J
2. If a refrigerator is a heat pump that follows the first law of
thermodynamics, how much heat was removed from food inside of the
refrigerator if it released 492J of energy to the room? Joules
The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room
=> Answer = 492 J
3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules
Formula: heat = mass * specific heat * ΔT
specific heat of water = 4186 J / Kg °C
heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
</span>