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3 years ago

8

your friend has an electric wheelchair that can travel at a speed of 352 feet per minute. how far can your friend travel in 15 m

inute?

2 answers:

Oksi-84 [34.3K]3 years ago

8 0

352x15=5,280????? I don't know if I'm right. Sorry.

Arada [10]3 years ago

5 0

Eh can travel 4875 feet in 15 minutes.
Hope this helps!

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3 years ago

Jodi poured herself a cold soda that had an initial temperature 36 degrees F and immediately went outside to sunbathe where the

mr Goodwill [35]

this can be solve using newtons heating of cooling

(Ts – T) =(Ts – To)*e^(-kt)

Where Ts is the ambient temperature

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3 years ago

A simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week

Levart [38]

Answer:

99% confidence interval for the population proportion of employed individuals is [0.104 , 0.224].

Step-by-step explanation:

We are given that a simple random sample of size n=250 individuals who are currently employed is asked if they work at home at least once per week.

Of the 250 employed individuals surveyed, 41 responded that they did work at home at least once per week.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of individuals who work at home at least once per week = $\frac{41}{250}$ = 0.164

n = sample of individuals surveyed = 250

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<em>99% confidence interval for p</em> = [ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.164-2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } }$ , $0.164+2.5758 \times {\sqrt{\frac{0.164(1-0.164)}{250} } }$ ]

= [0.104 , 0.224]

Therefore, 99% confidence interval for the population proportion of employed individuals who work at home at least once per week is [0.104 , 0.224].

7 0

3 years ago

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Furkat [3]

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