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never [62]
3 years ago
14

Which represents if x=4 then y=-2

Mathematics
1 answer:
Kruka [31]3 years ago
7 0

Answer:

what are the answers?

Step-by-step explanation:

Otherwise it isn't answerable

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Find the x-value for point C such that AC and BC from a 2:3 ratio.
Galina-37 [17]

Given:

Point C divides AB such that AC:BC=2:3.

To find:

The x-value for point C.

Solution:

Section formula: If a point divide a line segment in m:n, then

Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)

Form the given graph it is clear that the endpoints of the line segment AB are A(-3,5) and B(3,0).

Point C divides AB such that AC:BC=2:3. Using section formula, the coordinates of point C are

C=\left(\dfrac{2(3)+3(-3)}{2+3},\dfrac{2(0)+3(5)}{2+3}\right)

C=\left(\dfrac{6-9}{5},\dfrac{0+15}{5}\right)

C=\left(\dfrac{-3}{5},\dfrac{15}{5}\right)

C=\left(-0.6,3\right)

The x-value of C is -0.6.

Therefore, the correct option is B.

6 0
3 years ago
An instructor at a major research university occasionally teaches summer session and notices that that there are often students
Mademuasel [1]

Answer:

Check the explanations

Step-by-step explanation:

According to given information, an instructor at a major research university occasionally teaches summer session and notices that that there are often students repeating the class. On the first day of class, she counts 105 students enrolled, of which 19 are repeating the class. The university enrolls 15,000 students.

Therefore the number of observation n 105 and enrolled x=19

1. An estimate of the population proportion repeating the class is given by:

c) 0.181.

Explanation:

19 --0.18090,181 n 105

2. The instructor wishes to estimate the proportion of students across campus who repeat a course during summer sessions and decides to do so on the basis of this class. Would you advise the instructor against it and why:

d) Yes, because this class is not a random sample of students.

3. The standard error for the estimated sample proportion is given by:

c) 0.0376.

Explanation:

SE  (\hat{p})=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}

0.181 × (1-0.181) 105

0,0376 0.0376

4. A 95% confidence interval is given by:

c) 0.107, 0.255.

Explanation:

In order to determine the 95% confidence interval we follow the following step:

Where the z value is determined from the standard normal table as ~ 1.96

\hat{p}\pm \left [z\times SE(\hat{p}) \right ]

0.181 ± (1.96 × 0.0376)

0.181 ± 0.0737

Therefore the lower confidence interval is

LCI= 0.181- 0.0737

LCI= 0.107

Therefore the upper confidence interval is

UCI = 0.181 + 0.0737

UCI0.255

Therefore 95% confidence interval is

\left (0.107,0.255 \right )

5. She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are:

d)0 :P = 0.10  and  H_{a}:p\neq 0.10

Explanation:

She hypothesizes that, in general, 10% of students repeat a course. The hypotheses to be tested are defined on the basis of observation,

Null hypothesis as:

H_{0}:p= 0.10

and alternative hypothesis as:

H_{a}:p\neq 0.10

6. The test statistic for this hypothesis is given by:

c) 2.765.

Explanation:

In order to determine the z test statistics as:

Z=\frac{\hat{p}-p_{0}}{\sqrt{\frac{p_{0}(1-p_{0})}{n}}}

 0.181 - 0.10 0.10x (1-0.10)

 =2.765

7. The P value for this test is:

c) 0.01>P>0.005 .

Explanation:

P value is calculated as:

P(Z > 2.765) 0.002845 for one tail test and

P(Z > 2.765) 0.005692 for two tail test.

8. Based on the p-value found:

b) we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.

Explanation:

As the z observed is more than the tabulated z value at 95% as:

Z_{observed}=2.765> Z_{tabulated}=1.96

and also P value is less than the \alpha =1-0.95=0.05

P(Z\geq 2.765)=0.005692< \alpha =0.05

Therefore we accept the alternative hypothesis and we may conclude that we have strong evidence that the proportion of students repeating a class during summer sessions is not 10%.

7 0
3 years ago
Help with question 9 please .
Andrei [34K]

Answer:

I wanna say that it's A. curved lines.

6 0
3 years ago
Tracy deposited $59 into a bank account that earned 3.5% simple interest each year. If no money was deposited into or withdrawn
Illusion [34]

Answer

Find out the how much money was in the account after 1\frac{1}{2} years.

To prove

Formula

Simple\ interest = \frac{Principle\times Rate\times time}{100}

As given

Tracy deposited $59 into a bank account that earned 3.5% simple interest each year.

Principle = $59

Rate = 3.5 %

Time = 1\frac{1}{2}

i.e

Time = \frac{3}{2}

Time = 1.5 years

Put all the values in the equation

Simple\ interest = \frac{59\times 3.5\times 1.5}{100}

                                 =  \frac{309.75}{100}[/tex]

                                = $3.1(approx)

Amount in the account = Principle + Simple interest

                                      = $59 + $ 3.1

                                      = $ 62.1

1 Dollar = 100 cent

Now convert $62.1 into cent.

= 62.1 × 100

= 6210 cent

Therefore the money in the account is 6210 cent .





8 0
4 years ago
Read 2 more answers
Carter ran 7/8 mile in a race. Taylor ran 1/4 mile. How much further for carter run than Taylor?
Artist 52 [7]

Answer:

1 17/20 is the answer

3 0
3 years ago
Read 2 more answers
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