All categories

3 years ago

HELP ASAP!!!!! 25 POINTS!!!!!!!! AND WILL GVE BRAINLIEST!!!!!!!!!

Mathematics

1 answer:

Volgvan3 years ago

7 0

• Im sorry but we doesnt know the x number :( but dont worry to plot the diagram:

- substitute the number x (given) into the equation f(x)

- example: f(x) = x + 9 / x - 3

- then x given is 4

- therefore, f(x) = (4) + 9 / (4) - 3

- f(x) = 13 / 1 or 13

- then you plot at 13 on the graph :)

hope this helps ;/

You might be interested in

What is the correct factored for of x^2-9/x-3 to show a common factor can be divided out?

natali 33 [55]

Answer:

should be x^2-9/x-3

3 0

3 years ago

Today, everything at a store is on sale. The store offers a 20% discount. The regular price of a shirt is $14. What is the disco

USPshnik [31]

Answer:

$11.20

Step-by-step explanation:

The regular price of the T-Shirt is $14 dollars. The discount is 20% off.

20 percent of 14:

14 * 0.2 = 2.80

You would save $2.80.

14 - 2.80 = 11.20

The discount price should be $11.20.

Hope this helps.

4 0

3 years ago

Solve. x² – 5x + 6 = 0 {–2, –3} {2, 3} {3, –2} {5, –1}

ddd [48]

What 2 numbes multiply go get 6 and add to get -5
-6 and 1

(x-6)(x+1)=0
set to zero

x-6=0
x=6

x+1=0
x=-1

last one

4 0

3 years ago

A television Mark 1600 was sold for 1200 what percent discount was given on that television

adoni [48]

Answer:

25%

Step-by-step explanation:

Discount = 1600 - 1200 = 400

Percentage discount

$= \frac{400}{1600} \times 100 \\ \\ = \frac{400}{16} \\ \\ = 25\%$

5 0

3 years ago

A random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used. A survey of 500 randomly selecte

vodka [1.7K]

Answer:

We conclude that the mean Ohio score is below the national average.

Step-by-step explanation:

We are given that a random survey of 1000 students nationwide showed a mean ACT score of 21.1. Ohio was not used.

A survey of 500 randomly selected Ohio scores showed a mean of 20.8. The population standard deviation is 3.

Let $\mu$ = mean Ohio scores.

So, Null Hypothesis, $H_0$ : $\mu$ $\geq$ 21.1 {means that the mean Ohio score is above or equal the national average}

Alternate Hypothesis, $H_A$ : $\mu$ < 21.1 {means that the mean Ohio score is below the national average}

The test statistics that would be used here One-sample z test statistics as we know about population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}$ ~ N(0,1)

where, $\bar X$ = sample mean Ohio score = 20.8

$\sigma$ = population standard deviation = 3

n = sample of Ohio = 500

So, test statistics = $\frac{20.8-21.1}{\frac{3}{\sqrt{500}}}$

= -2.24

The value of z test statistics is -2.24.

Now, at 0.1 significance level the z table gives critical value of -1.2816 for left-tailed test. Since our test statistics is less than the critical values of z as -2.24 < 1.2816, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean Ohio score is below the national average.

4 0

3 years ago