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Dennis_Churaev [7]

3 years ago

6

The Science Club went on a two-day field trip. The first day the members paid $30 for transportation plus $10 per ticket to the

planetarium. The second day they paid $85 for transportation plus $7 per ticket to the geology museum. Write an expression to represent the total cost in dollars for two days for the n members of the club.
emergency help mate i need answers now

1 answer:

Anettt [7]3 years ago

6 0

Answer: 132

Step-by-step explanation: For day A the students payed 40 dollars and on day B they paid 92 so 40+92=$132 in all.

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20pts Please help! Asap Visitor attendance at the nature center in December is one-quarter the attendance in September. In Decem

Leya [2.2K]

Answer: I am in 6th grade and I just did this test too at K12 CAVA. I got the answer correct, it is:

376

Hope this helps.

3 0

3 years ago

Lori is picking up lunch for herself and her coworkers. Some of her coworkers ordered the salad which costs $7.50, and some of h

Step2247 [10]

Answer: 3 people ordered soup.

Step-by-step explanation:

Let x be the number of people ordered salad and y be the number of people ordered soup.

As per given,

x=2y (i)

7.50x+6.25y= 63.75 (ii)

Substitute value of c =2y in (ii) , we get

7.50(2y)+6.25y= 63.75

⇒ 15y+6.25y = 63.75

⇒21.25y = 63.75

⇒ y= 3 [divide both sides by 21.25]

Then, x= 2(3) =6

So 6 people ordered salad and 3 people ordered soup.

4 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Contact [7]

Answer:

(A) A 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval would increase.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval would decrease.

(D) A 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

Step-by-step explanation:

We are given that 19 students are randomly selected the sample mean was $390 and the standard deviation was $120.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = $390

s = sample standard deviation = $120

n = sample of students = 19

$\mu$ = population mean

<em>Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation. </em>

<u>So, 95% confidence interval for the population mean, </u> $\mu$ <u> is ; </u>

P(-2.101 < $t_1_8$ < 2.101) = 0.95 {As the critical value of t at 18 degrees of

freedom are -2.101 & 2.101 with P = 2.5%}

P(-2.101 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.101) = 0.95

P( $-2.101 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

<u> 95% confidence interval for</u> $\mu$ = [ $\bar X-2.101 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.101 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $\$390-2.101 \times {\frac{\$120}{\sqrt{19} } }$ , $\$390+2.101 \times {\frac{\$120}{\sqrt{19} } }$ ]

= [$332.16, $447.84]

(A) Therefore, a 95% confidence for the population mean is [$332.16, $447.84] .

(B) If the confidence level in part (a) changed from 95% to 99%, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would increase because of an increase in the z value.

(C) If the sample size in part (a) changed from 19 to 22, then the margin of error for the confidence interval which is $Z_(_\frac{\alpha}{2}_) \times \frac{s}{\sqrt{n} }$ would decrease because as denominator increases; the whole fraction decreases.

(D) We are given that to estimate the proportion of students who purchase their textbooks used, 500 students were sampled. 210 of these students purchased used textbooks.

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 students who purchase their used textbooks = $\frac{210}{500}$ = 0.42

n = sample of students = 500

p = population proportion

<em>Here for constructing a 99% confidence interval we have used a One-sample z-test statistics for proportions</em>

<u>So, 99% confidence interval for the population proportion, p is ; </u>

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5%

level of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

<u> 99% confidence interval for</u> p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.42 -2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ , $0.42 +2.58 \times {\sqrt{\frac{0.42(1-0.42)}{500} } }$ ]

= [0.363, 0.477]

Therefore, a 99% confidence interval for the proportion of students who purchase used textbooks is [0.363, 0.477] .

8 0

3 years ago

11x² + 2x + 16<br> 2x² + x + 2<br> -

Anit [1.1K]

Answer: 13x^(2)+3x+18

Step-by-step explanation:

13x^(2)+3x+18

4 0

4 years ago

A baby pool is shaped like a short, wide cylinder and holds 52 ft3 of water. A medium pool measures 4 times the height and width

otez555 [7]

The volume of a cylinder is the area of the base, times the height.
<span>The area of the circular base is, of course, π times the square of its radius. </span>
<span>So the volume of a cylinder, V, is </span>
<span>V = π h r^2 </span>
<span>where h is the height and r is the radius. </span>

<span>If we multiply the height by 3, the new height is 3h. </span>
<span>If we multiply the width by 3, that multiplies the radius (which is half the width) by 3 also. </span>
<span>So the new volume would be </span>
<span>π (3h) (3r)^2 = π (3h) (9r^2) = 27π h r^2 </span>
<span>which is 27 times π h r^2, the volume of the smaller pool. But we know that volume is 37 cubic feet. </span>

<span>So the volume of the larger pool is </span>
<span>27 * 37 cubic feet = 999 cubic feet [answer B] </span>

<span>This is an instance of a general rule: </span>
<span>If we multiply ALL the linear dimensions of an object by a factor F, </span>
<span>while keeping the shape the same (often termed "similar" in geometry), </span>
<span>that multiplies the volume by F^3 (the cube of F). </span>

<span>It also multiplies the surface area by F^2. </span>

<span>These rules are very useful, especially when you hit these questions on an exam under time constraints. If you know them, you'll save time you can use to work other questions.</span>

3 0

3 years ago