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erastova [34]
3 years ago
11

A survey was given to a random sample of students attending college in which they were asked, "As a college student, do you feel

like your high school prepared you for your college experience?" The data showed that 54% of the 600 students surveyed replied yes. What is the population of the statistical question? A. All students attending college B. All students in high school and college C. The 600 students who were surveyed D. The 54% of students who replied yes​
Mathematics
1 answer:
Phantasy [73]3 years ago
3 0

Answer:

D

Step-by-step explanation:

Option A doesn't give any statistical data, B is the same as A, and C doesn't show accurately how many students replied yes or no.

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2.2
Ludmilka [50]

Answer:

R180

Step-by-step explanation:

Adjusted income = (original pocket money x new ratio) / old ratio

( 6 x $150) / 5 = $180

4 0
3 years ago
Can someone please help me with this?
elena-14-01-66 [18.8K]

Answer:

y=2/5x+2

slope: 2/5

y=-1/4x+4

y-intercept: 4

Step-by-step explanation

The equation for slope-intercept form is y=mx+b.

m=the slope

b= the y-intercept

5 0
3 years ago
The area of the triangle formed by x− and y− intercepts of the parabola y=0.5(x−3)(x+k) is equal to 1.5 square units. Find all p
Juliette [100K]

Check the picture below.


based on the equation, if we set y = 0, we'd end up with 0 = 0.5(x-3)(x-k).

and that will give us two x-intercepts, at x = 3 and x = k.

since the triangle is made by the x-intercepts and y-intercepts, then the parabola most likely has another x-intercept on the negative side of the x-axis, as you see in the picture, so chances are "k" is a negative value.

now, notice the picture, those intercepts make a triangle with a base = 3 + k, and height = y, where "y" is on the negative side.

let's find the y-intercept by setting x = 0 now,


\bf y=0.5(x-3)(x+k)\implies y=\cfrac{1}{2}(x-3)(x+k)\implies \stackrel{\textit{setting x = 0}}{y=\cfrac{1}{2}(0-3)(0+k)} \\\\\\ y=\cfrac{1}{2}(-3)(k)\implies \boxed{y=-\cfrac{3k}{2}} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of a triangle}}{A=\cfrac{1}{2}bh}~~ \begin{cases} b=3+k\\ h=y\\ \quad -\frac{3k}{2}\\ A=1.5\\ \qquad \frac{3}{2} \end{cases}\implies \cfrac{3}{2}=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)


\bf \cfrac{3}{2}=\cfrac{3+k}{2}\left( -\cfrac{3k}{2} \right)\implies \stackrel{\textit{multiplying by }\stackrel{LCD}{2}}{3=\cfrac{(3+k)(-3k)}{2}}\implies 6=-9k-3k^2 \\\\\\ 6=-3(3k+k^2)\implies \cfrac{6}{-3}=3k+k^2\implies -2=3k+k^2 \\\\\\ 0=k^2+3k+2\implies 0=(k+2)(k+1)\implies k= \begin{cases} -2\\ -1 \end{cases}


now, we can plug those values on A = (1/2)bh,


\bf \stackrel{\textit{using k = -2}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-2)\left(-\cfrac{3(-2)}{2} \right)\implies A=\cfrac{1}{2}(1)(3) \\\\\\ A=\cfrac{3}{2}\implies A=1.5 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{using k = -1}}{A=\cfrac{1}{2}(3+k)\left(-\cfrac{3k}{2} \right)}\implies A=\cfrac{1}{2}(3-1)\left(-\cfrac{3(-1)}{2} \right) \\\\\\ A=\cfrac{1}{2}(2)\left( \cfrac{3}{2} \right)\implies A=\cfrac{3}{2}\implies A=1.5

7 0
3 years ago
Ms. Carson's art class made a large rectangular mural. The length of the mural is 9 feet, and it's area is 72 square feet. What
lorasvet [3.4K]
You do the reverse and divide the 2 values to find out the last one

72/9 = 8

Do you get it?
8 0
3 years ago
Read 2 more answers
Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. ​(a) Descri
OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

\mu = 74

and

\sigma = 6

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

\mu = 74

The standard error of the means becomes the standard deviation of the sampling distribution.

\sigma_ { \bar X }  =  \frac{ \sigma}{ \sqrt{n} }  \\ \sigma_ { \bar X }  =  \frac{ 6}{ \sqrt{36} }  = 1

Part b) We want to find

P(\bar X \:>\:75.9)

We need to convert to z-score.

P(\bar X \:>\:75.9)  = P(z \:>\: \frac{75.9 - 74}{1} )  \\  = P(z \:>\: \frac{75.9 - 74}{1} ) \\  = P(z \:>\: 1.9) \\  = 0.0287

Part c)

We want to find

P(\bar X \: < \:71.95)

We convert to z-score and use the normal distribution table to find the corresponding area.

P(\bar X \: < \:71.95)  = P(z \: < \: \frac{71.9 5- 74}{1} )  \\  = P(z \: < \: \frac{71.9 5- 74}{1} ) \\  = P(z \: < \:  - 2.05) \\  = 0.0202

Part d)

We want to find :

P(73\:

We convert to z-scores and again use the standard normal distribution table.

P( \frac{73 - 74}{1} \:< \: z

5 0
3 years ago
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