Let y = 2t + 5 be a linear function representing the distance from home for an ant t minutes after starting out from a location
2 answers:
Answer:
The correct option is B.
Step-by-step explanation:
The given equation is
..... (1)
Where, y is distance from home for an ant t minutes after starting out from a location near its home.
The slope intercept form of a line is
.... (2)
Where, m is slope and b is y-intercept or initial value.
On comparing (1) and (2), we get
Here, 5 represent the initial distance of ant from its home.
It means the ant is moving at 2 feet per minute and the ant started out 5 feet from its home.
Therefore option B is correct.
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Answer:
95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].
Step-by-step explanation:
We are given that in a certain rural county, a public health researcher spoke with 111 residents 65-years or older, and 28 of them had obtained a flu shot.
Firstly, the Pivotal quantity for 95% confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of residents 65-years or older who had obtained a flu shot =
= 0.25
n = sample of residents 65-years or older = 111
p = population proportion of residents who were getting the flu shot
<em>Here for constructing 95% confidence interval we have used One-sample z test for proportions.</em>
<u>So, 95% confidence interval for the population proportion, p is ;</u>
P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level
of significance are -1.96 & 1.96}
P(-1.96 < < 1.96) = 0.95
P( <
<
) = 0.95
P( < p <
) = 0.95
<u>95% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.169 , 0.331]
Therefore, 95% confidence interval for the percent of the 65-plus population that were getting the flu shot is [0.169 , 0.331].