MATHEMATICS HIGH SCHOOL

What is the exact volume of the cone 4 in and 7 in

Answers

Answer 1

Answer:

Answer: V=205.25 cubic inches.

Step-by-step explanation:

For the given cone height = 4 in and radius= 7 in .Volume of any cone can be calculated by the formula :

V= $(1)/(3) \pi r^(2) h$

Substituting the values of r and h we have:

V= $(1)/(3) \pi 7^(2) .4$

Solving for V we have:

V=205.25 cubic inches.

Answer 2

Answer: V = (1/3) *pi* (7)^2 * 4

V = ( 196 pi ) / 3

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The surface area of any shape is expressed in?a. feet and inches.
b. square units.
c. cubic units.
d. length and width.

Answers

the surface of any shape often is a result of multiple two dimension so it expressed in square units
the answer is B

b) square units is the answer

On a map, 1/3 inch equals 15 miles. The distance between two towns on a map is 3 2/3 inches. How many miles are actually between the two towns?a. 11
b. 16
c. 88
d. 132
e. 165

Answers

15x11=165 E is the answer.

Simplify
-2(3x - 4y) +
5(x+3y)

Answers

Answer:

$- x + 23y$

Step-by-step explanation:

$- 2(3x - 4y) + 5(x + 3y) \n - 6x + 8y + 5x + 15y \n - 6x + 5x - 8y + 23y \n = - x + 7y$

Answer:

-x + 23y

Step-by-step explanation:

-2(3x - 4y) + 5(x+3y)

Expand the brackets.

-6x + 8y + 5x + 15y

Rearrange.

- 6x + 5x +8y + 15y

Add or subtract like terms.

- 1x + 23y

A cube has a side length of 12 meters. What is the volume of the cube? A. 36 m3 B. 144 m3 C. 864 m3 D. 1728 m3

Answers

$The\ volume\ of\ the\ cube:V=a^3\ \ /a-the\ length\ the\ edge\ of\ the\ cube\n\na=12\ m\n\ntherefore\n\n\boxed{V=12^3= 1728\ (m^3)}$

A cube has congruent sides therefore the volume is 12*12*12=1728 meters cubed

Among the 10 most popular sports, men include competition-type sports - pool and billiards, basketball, and softball - whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of fishing, with 41% of those surveyed indicating that they had fished during the year. Suppose 180 randomly selected men are asked whether they had fished in the past year. Suppose 180 randomly selected men are asked whether they has fished in the past year.a. What is the probability that fewer than 50 had fished?
b. What is the probability that between 50 and 75 (inclusive) had fished?
c. If the 180 men selected for the interview were selected by the marketing department of a sporting-goods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results?

Answers

Answer:

(a) The probability that fewer than 50 had fished is 0.0002.

(b) The probability that between 50 and 75 (inclusive) had fished is 0.6026.

Step-by-step explanation:

Let X = number of men who had fished during the year.

The probability of the random variable X is, p = 0.41.

A random sample of n = 180 men are selected.

The random variable X follows a Binomial distribution with parameters n = 180 and p = 0.41.

A Normal approximation to Binomial is applied when the following conditions are met,

np ≥ 10
n(1 - p) ≥ 10

Check:

$np=180*0.41=73.8>10\nn(1-p)=180* (1-0.41)=63.72>10$

Thus, the distribution of x can be approximated by a Normal distribution with:

Mean = $np=180*0.41=73.8$

Standard deviation = $√(np(1-p))=√(180*0.41*(1-0.41))=6.599$

(a)

Compute the probability that fewer than 50 had fished as follows:

$P(X<50)=P((X-\mu)/(\sigma)>(50-73.8)/(6.599))\n=P(Z<-3.61)\n=1-P(Z<3.61)\n=1-0.9998\n=0.0002$

Thus, the probability that fewer than 50 had fished is 0.0002.

(b)

Compute the probability that between 50 and 75 (inclusive) had fished as follows:

$P(50\leq X\leq 75)=P(49.5<X<75.5)\n=P((49.5-73.8)/(6.599)<(X-\mu)/(\sigma)<(75.5-73.8)/(6.599))\n=P(-3.68<Z<0.26)\n=P(Z<0.26)-P(Z<-3.68)\n=0.6026-0\n=0.6026$

Thus, the probability that between 50 and 75 (inclusive) had fished is 0.6026.

(c)

If the sample of 810 men are selected from mailing list then it is highly probable that the sample is not a representative of the true population, i.e. sports men.

Because the people interested in sports are less likely to be interested in fishing.

Thus, it could be concluded that the survey results are not reliable.

Write a fraction that is a multiple of 4/5.

Answers

so if we assume that the ending fraction has to be equal to the origonal fraction then we multiply it by 1 or x/x where x=x

so 4/5 times 2/2=8/10
4/5 times 3/3=12/15
and so on

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