Answer:
y= -7x + 1
Step-by-step explanation:
this is the answer
Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
In 2015, eighty-seven babies per one thousand died before reaching one year of age. This figure is called the infant mortality rate.
Infant mortality is defined as a child dying before turning one. The number of newborn deaths for every 1,000 live births is known as the infant mortality rate. The infant mortality rate is a significant indicator of the general health of a society in addition to providing us with valuable information on maternal and baby health. In the United States, there were 5.4 infant deaths for every 1,000 live births in 2020.
Learn more about infant mortality rate here: brainly.com/question/13493110
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Answer:
idk let me go look at it
Step-by-step explanation:
