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2 years ago

Solve the system! - 3x + 3y = 4 y=x+3 What is the solution?

Mathematics

1 answer:

STatiana [176]2 years ago

3 0

Answer:

This system has no solution

Step-by-step explanation:

If this is the real question,it has no solution

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Will mark as Brainliest.

AfilCa [17]

C) is the right answer :)))))

8 0

4 years ago

Read 2 more answers

Lace edging is sold by the foot for $0.80 or by the foot for $1.99. Which is more expensive

vampirchik [111]

Answer:

$1.99 is more expensive.

Step-by-step explanation:

Given:

One type of lace edging cost $0.80 per foot

Another type of lace edging cost $1`.99 per foot.

We need to find which is more expensive.

In $0.88 where one's place is 0.

In $1.99 where one's place is 1.

By comparing the one's place obviously $1.99 more expensive.

Hope you will understand the concept.

Thank you.

6 0

3 years ago

How do you solve this

Dmitriy789 [7]

By definition of the trig functions,

$\sin30^\circ=\dfrac y{16}$

$\cos30^\circ=\dfrac x{16}$

We have $\sin30^\circ=\dfrac12$ , so $y=8$ , and $\cos30^\circ=\dfrac{\sqrt3}2$ , so $x=8\sqrt3$ .

Just to check: we should have

$x^2+y^2=16^2$

by the Pythagorean theorem. Indeed,

$8^2+(8\sqrt3)^2=4\cdot8^2=2^2(2^3)^2=2^8=(2^4)^2=16^2$

6 0

4 years ago

Can you help me solve this?

Wittaler [7]

Answer:

y=-1/2x+4 Is your answer

Step-by-step explanation:

y=mx+b

3=2x+4

-1=2x

x=-1/2

You already know y is 4 so y=-1/2x+4

4 0

3 years ago

UCF believes that the average time someone spends in the gym is 56 minutes. The university statistician takes a random sample of

11111nata11111 [884]

Answer:

We conclude that the average time someone spends in the gym is different from 56 minutes.

Step-by-step explanation:

We are given that UCF believes that the average time someone spends in the gym is 56 minutes.

The university statistician takes a random sample of 32 gym goers and finds the average time of the sample was 50 minutes. Assume it is known the standard deviation of time all people spend in the gym is 8 minutes.

Let $\mu$ = population average time someone spends in the gym

SO, Null Hypothesis, $H_0$ : $\mu$ = 56 minutes {means that the average time someone spends in the gym is 56 minutes}

Alternate Hypothesis, $H_A$ : $\mu\neq$ 56 minutes {means that the average time someone spends in the gym is different from 56 minutes}

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

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

where, $\bar X$ = sample average time someone takes in the gym = 50 min

$\sigma$ = population standard deviation = 8 minutes

n = sample of gym goers = 32

So, test statistics = $\frac{50-56}{\frac{8}{\sqrt{32} } }$

= -4.243

Since in the question we are not given the level of significance so we assume it to b 5%. Now at 5% significance level, the z table gives critical value between -1.96 and 1.96 for two-tailed test. Since our test statistics does not lie within the range of critical values of z so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the average time someone spends in the gym is different from 56 minutes.

4 0

3 years ago