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3 years ago

The volume of a cone is 800 cubic cm and the radius is 9 cm. Calculate the height.

Mathematics

1 answer:

scZoUnD [109]3 years ago

4 0

Answer:

Step-by-step explanation:

$v =\frac{1}{3} \pi r^{2} h$

$800=\pi 9^{2} /3$

$800*3=\pi 81 *height$

$height=2400/81 * \pi$

=2400/254.34

=9.4362

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2 years ago

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Answer:

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Step-by-step explanation:

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4 years ago

A researcher believes that 12% of a simple random sample of adults will be able to identify a Toyota Scion by brand and model na

DochEvi [55]

Answer: 650

Step-by-step explanation:

When prior estimate of population proportion is known , then the formula to find the required sample size is given by :-

$n=p(1-p)(\dfrac{z^*}{E})^2$

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E= margin of error

z* = Critical value.

Let p be the proportion of adults able to identify a Toyota Scion by brand and model name.

As per given , we have

p = 12%= 0.12

E= 2.5%=0.025

Critical value for 95% confidence interval : z* = 1.960 [By z-table ]

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3 years ago

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

Leviafan [203]

Answer:

Probability of having student's score between 505 and 515 is 0.36

Given that z-scores are rounded to two decimals using Standard Normal Distribution Table

Step-by-step explanation:

As we know from normal distribution: z(x) = (x - Mu)/SD

where x = targeted value; Mu = Mean of Normal Distribution; SD = Standard Deviation of Normal Distribution

Therefore using given data: Mu (Mean) = 510, SD = 10.4 we have z(x) by using z(x) = (x - Mu)/SD as under:

In our case, we have x = 505 & 515

Approach 1 using Standard Normal Distribution Table:

z for x=505: z(505) = (505-510)/10.4 gives us z(505) = -0.48

z for x=515: z(515) = (515-510)/10.4 gives us z(515) = 0.48

Afterwards using Normal Distribution Tables and rounding the values to two decimals we find the probabilities as under:

P(505) using z(505) = 0.32

Similarly we have:

P(515) using z(515) = 0.68

Now we may find the probability of student's score between 505 and 515 using:

P(505 < x < 515) = P(515)-P(505) = 0.68 - 0.32 = 0.36

PS: The standard normal distribution table is being attached for reference.

Approach 2 using Excel or Google Sheets:

P(x) = norm.dist(x,Mean,SD,Commutative)

P(505) = norm.dist(505,510,10.4,1)

P(515) = norm.dist(515,510,10.4,1)

Probability of student's score between 505 and 515= P(515) - P(505) = 0.36

Download pdf

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