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3 years ago

14

a company makes a cone-shaped container with a height of 15 in. The area of its base is about 78.8 in.2 Approximately what is th

e volume if the container?

1 answer:

Pepsi [2]3 years ago

6 0

Answer:

Approximately the volume of cone-shaped container is 393 in³.

Step-by-step explanation:

You might be interested in

A scientist found that x grams of Metal A is completely oxidized in 2x13 seconds and x grams of Metal B is completely oxidized i

ExtremeBDS [4]

Answer: Metal A is 3 times faster than Metal B.

Step-by-step explanation:

Since we have given that

Time taken to oxidised metal A = $2x\sqrt{3}\ seconds$

Time taken to oxidised metal B = $6x\sqrt{3}$ seconds

So, ratio of metal A to metal B would be

$\dfrac{Metal\ A}{Metal\ B}=\dfrac{2x\sqrt{3}}{6x\sqrt{3}}=\dfrac{1}{3}$

So, Ratio of Metal A to Metal B is 1 : 3.

Hence, Metal A is 3 times faster than Metal B.

4 0

3 years ago

ehidna [41]

Yesterday, 1/4 of 80 =20 gave their speeches, so there are 80-20=60 students left.
Today, 2/5 of 60=24 students gave their speeches, so there are 60-24=36 students who still haven't give their speeches
another way to think of it is 80*(1- 1/4)*(1- 2/5)=80*(3/4)*(3/5)=36

4 0

3 years ago

An OSU senior is studying for exams in psychology and economics. The student has time to read 50 pages of psychology and 10 page

cestrela7 [59]

Answer:

3 Pages

Step-by-step explanation:

Let the pages of economics read = e
Let the pages of psychology read = p
Let the total time taken on each instance=t

In the first instance, the student has time to read 50 pages of psychology and 10 pages of economics.

t=50p+10e

The student could read 30 pages of psychology and 70 pages of economics.

t=30p+70e

Since the two situations take the same amount of time, we have:

50p+10e=30p+70e

Collect like terms

50p-30p=70e-10e

20p=60e

Divide both sides by 20

p=3e

Therefore, in the time it will take the student to read 1 page of psychology, the student can read 3 pages of economics.

8 0

3 years ago

Use the net as an aid to compute the surface area of the triangular prism. A) 108 cm2 B) 120 cm2 C) 132 cm2 D) 170 cm2

nadya68 [22]

Step-by-step explanation:

Since there is no diagram of the triangular prism that we are required to calculate it's surface area. We can just fix random dimensions in order to enable us understand what we are required to do.

Meanwhile, a triangular prism is a solid shape that possesses two parallel triangles or triangular faces and about three rectangles or rectangular faces.

Example 1:

Calculate the surface area of a triangular prism with side 8cm, base 6cm and height 7cm.

Formula for calculating the surface area of a triangular prism is;

ab + 3bh

Where a = side of the prism

b = base of one of the triangular faces

h = height not the triangular face/prism.

Here a = 8cm

b = 6cm

c = 7cm

Substituting properly;

(8 × 6) + 3(6 × 7)

= 48 + 126

= 174cm^2. The surface area of the triangular prism is 174cm^2.

Example 2.

A triangular prism with side 9cm, height 4cm and base 5cm will be wrapped with a sheet of wrapping foil. Calculate the surface area of this rectangular prism.

The formula for calculating the surface area of a triangular prism = ab + 3bh

Where a = side

b = base

h = height of prism

In this case, a = 9cm

b = 5cm

h = 4cm

Substituting appropriately, we will have:

(9 × 5) + 3(5 × 4)

45 + 60

= 105cm^2

So, the surface area of this triangular prism is 105cm^2

8 0

3 years ago

Read 2 more answers

g In R simulate a sample of size 20 from a normal distribution with mean µ = 50 and standard deviation σ = 6. Hint: Use rnorm(20

Illusion [34]

Answer:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

Step-by-step explanation:

For this case first we need to create the sample of size 20 for the following distribution:

$X\sim N(\mu = 50, \sigma =6)$

And we can use the following code: rnorm(20,50,6) and we got this output:

> a<-rnorm(20,50,6)

> a

[1] 51.72213 53.09989 59.89221 32.44023 47.59386 33.59892 47.26718 55.61510 47.95505 48.19296 54.46905

[12] 45.78072 57.30045 57.91624 50.83297 52.61790 62.07713 53.75661 49.34651 53.01501

Then we can find the mean and the standard deviation with the following formulas:

> mean(a)

[1] 50.72451

> sqrt(var(a))

[1] 7.470221

5 0

3 years ago