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Find volume with full working out pls asap

aev [14]

<h3>Answer: approximately 2076.56001909938 cubic cm</h3>

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Work Shown:

V1 = volume of cylinder

V2 = volume of cone on top

V3 = V1+V2 = volume of entire 3D solid figure

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V1 = volume of cylinder

V1 = pi*r^2*h

V1 = pi*8.75^2*5.8

V1 = 1395.06348773471 which is approximate

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V2 = volume of cone

V2 = (1/3)*pi*r^2*h

V2 = (1/3)*pi*8.75^2*8.5

V2 = 681.49653136466 which is approximate

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V3 = V1+V2

V3 = 1395.06348773471+681.49653136466

V3 = 2076.56001909938

Answer is approximate

The units for the volume are in cubic cm.

5 0

3 years ago

In an arithmetic sequence, if you are given a3=93 and a5=135, find a=100. Must show work to receive full credit.

Elanso [62]

The 100th term of the sequence is 2130.

<h3>How to calculate the value?</h3>

a3 = a + 2d = 93

a5 = a + 4d = 135

Compute both equations

(4d - 2d) = (135 - 93)

2d = 42

d = 42/2 = 21

a + 2d = 93

a + 2(21) = 93

a + 42 = 93

a = 94 - 42 = 51

100th term will be:

= a + 99d

= 51 + 99(21)

= 51 + 2079

= 2130

Learn more about sequence on:

brainly.com/question/6561461

#SPJ1

5 0

1 year ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

So

$P(x \ge 60) \approx P(z > 1.34)$