Solve the equation -10 (g

A forest ranger wants to install a 1,500 CFM (cubic feet per minute) fan. A ranger station consists of a rectangular prism and a

levacccp [35]

Sample response: Yes, the fan will provide enough airflow. The volume of the ranger station is 720 cubic feet. The ranger station requires a 720 CFM fan. 1,500 CFM is greater than 720 CFM, so it will be enough airflow.

6 0

2 years ago

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Your friend really wants to buy the newest, coolest cell phone which will cost him $310. He gets a job detailing cars and SUVs a

LuckyWell [14K]

There is multiple ways but one of them is 3 SUV and 1 car and that will give him $315.

I hope this helps :)

5 0

2 years ago

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Workers have packed 1,400 glasses in 7 boxes. If they pack 3 more boxes, how many glasses will they have packed in all?

Alexandra [31]

2000

1box = 1400/7 = 200
200×3=600
1400+600=2000

Step-by-step explanation:
Given :Workers have packed 1,400 glasses in 7 boxes.
To Find :If they pack 3 more boxes, how many glasses will they have packed in all?
Solution:
Workers packed no. of glasses in 7 boxes = 1400
Workers packed no. of glasses in 1 box =
Workers packed no. of glasses in 3 boxes =
=
So, initially they packed 1400 glasses
If they pack 3 more boxes so, the pack 600 glasses more
So, The total no. of glasses have packed by workers = 1400+600 = 2000
Hence they have packed 2000 glasses in all.

5 0

2 years ago

In a country where prices are rising quickly, bread that now costs $1.39 will cost 2.4 times as much next year. How much will th

Lynna [10]

Answer:

$3.34

Step-by-step explanation:

$1.39 x 2.4 = $3.336

4 0

2 years ago

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Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the 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 Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$