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4 years ago

6r-2(r+8)=11 Solve equation

1 answer:

4 0

$6r-2(r+8)=11\ \ \ \ |\text{use distributive property}\\\\6r-2(r)-2(8)=11\\\\6r-2r-16=11\ \ \ \ \ |+16\\\\4r=27\ \ \ \ \ |:4\\\\r=6.75$

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3 years ago

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A dog is chasing a cat towards a tree. The cat has a 10-yard lead and runs at a speed of 6 yards per second. The dog runs at a s

11111nata11111 [884]

Answer:

The dog will reach the tree first.

Step-by-step explanation:

If the tree is 30 yards from the dog's starting position, and the cat is 10 yards ahead of the dog, then the cat is 20 yards away from the tree. The dog runs at 8 yps, and the cat runs at 6yps. The cat would get there in approximately 3 seconds, and the dog would be ahead of the cat by 3 yards within 2 seconds, so the dog would be there before the cat would.

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3 years ago

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umka2103 [35]

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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.

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3 years ago

Amelia, Luis, Shauna, and Clarence used different approaches to solve the inequality

mezya [45]

Answer:

Luis’s, because he flipped the inequality sign when he subtracted

Step-by-step explanation:

7.2b + 6.5 > 4.8b – 8.1.

Amelia started by subtracting 7.2b from both sides to get

6.5 > –2.4b – 8.1

Correct

Luis started by subtracting 4.8b from both sides to get

2.4b + 6.5 < – 8.1

Incorrect

Shauna started by subtracting 6.5 from both sides to get

7.2b > 4.8b – 14.6

Correct

Clarence started by adding 8.1 to both sides to get

7.2b + 14.6 > 4.8b

Correct

Luis’s is incorrect because he flipped the inequality sign when he subtracted

6 0

3 years ago