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3 years ago

Item 9 Find the surface area of the cone with diameter d and slant height l. Round to the nearest tenth. d=12 cm l=85mm

Mathematics

1 answer:

PolarNik [594]3 years ago

5 0

Answer: $160.28\ cm^2$

Step-by-step explanation:

Given

diameter of cone $d=12\ cm$

radius is $r=\frac{d}{2}=6\ cm$

The slant height of the cone $l=85\ mm\approx 8.5\ cm$

The surface area of a cone is

$\Rightarrow A=\pi rl$

Substitute the value

$\Rightarrow A=\frac{22}{7}\times 6\times 8.5=160.28\ cm^2$

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8% of 70 is what number? (please explain)

Bingel [31]

Do a proportion
8/100= x/70
Cross multiply and divide
70 x 8= 560
560 divided by 100= 5.6
Answer is 5.6

You also could have done .08 x 70 which is also 5.6

7 0

3 years ago

Help please :)

Veronika [31]

Answer

500

Step-by-step explanation:

If the biologist tags 25 birds and lets them go, and then catches 100 and 5 of them have the tags, that would mean 20 of them are still out in the wild.

The way to solve this is if you put it into a proportion.

5/100 = 25/x

5x = (100)(25)

5x = 2500

5x/5 = 2500/5

x = 500

5/100=25/500

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3 years ago

Suppose that scores on the mathematics part of a test for eighth-grade students follow a Normal distribution with standard devia

riadik2000 [5.3K]

Answer:

We need an SRS of scores of at least 153.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How large an SRS of scores must you choose?

This is at least n, in which n is found when $M = 20, \sigma = 150$ . So

$M = z*\frac{\sigma}{\sqrt{n}}$

$20 = 1.645*\frac{150}{\sqrt{n}}$

$20\sqrt{n} = 1.645*150$

$\sqrt{n} = \frac{1.645*150}{20}$

$\sqrt{n} = 12.3375$

$(\sqrt{n})^{2} = (12.3375)^{2}$

$n = 152.2$

Rounding to the next whole number, 153

We need an SRS of scores of at least 153.

8 0

3 years ago

About how many times does a chickens heart beat in 1. Min

sammy [17]

Hello, a chickens heart beats 280-315 times per minute.

Happy Studies <3

8 0

3 years ago

Read 2 more answers

The service department of a luxury car dealership conducted research on the amount of time its service technicians spend on each

mart [117]

Answer:

Probability that the mean service time is between 1 and 2 hours is 0.96764.

Step-by-step explanation:

We are given that a systematic random sample of 100 service appointments has been collected.

The 100 appointments showed an average preparation time of 90 minutes with a standard deviation of 140 minutes.

Let $\bar X$ = sample mean service time

The z-score probability distribution for sample mean is given by;

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

where, $\mu$ = average preparation time = 90 minutes

$\sigma$ = standard deviation = 140 minutes

n = sample of appointments = 100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, probability that the mean service time is between 60 and 120 minutes is given by = P(60 minutes < $\bar X$ < 120 minutes)

P(60 minutes < $\bar X$ < 120 minutes) = P( $\bar X$ < 120 min) - P( $\bar X$ $\leq$ 60 min)

P( $\bar X$ < 120 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{120-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z < 2.14) = 0.98382

P( $\bar X$ $\leq$ 60 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{60-90}{\frac{140}{\sqrt{100} } }$ ) = P(Z $\leq$ -2.14) = 1 - P(Z < 2.14)

= 1 - 0.98382 = 0.01618

The above probability is calculated by looking at the value of x = 2.14 in the z table which has an area of 0.98382.

Therefore, P(60 min < $\bar X$ < 120 min) = 0.98382 - 0.01618 = 0.96764

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4 years ago