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3 years ago

The volume of a cube is 7.14 cubic meters. What is the cross-sectional area that is parallel to one of its faces?

Mathematics

1 answer:

Snowcat [4.5K]3 years ago

8 0

Answer:

A = 3.706 square meters

Step-by-step explanation:

the length of a cube has equal side.

therefore, the volume of a cube is given by S³

V = S³ = 7.14

where S is the length of a side

the surface area of a cube is = 6S²

where the area a aside is calculated, then it is multiply by 6.

if S³ = 7.14

S = ∛(7.14) = 1.925 m

What is the cross-sectional area that is parallel to one of its faces?

this is saying we should calculate for the cross-sectional area of one face. All faces in a cube is equal and parallel to each other.

the crossectional area of one side of a cube is = S² = 1.925² = 3.706 square meters

A = 3.706 square meters

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Type the correct answer in the box

ivolga24 [154]

Answer:

$w=V/(lh)$

Step-by-step explanation:

Let

l------> the length of the base of the prism

w------> the width of the base of the prism

h------> the height of the prism

we know that

The volume of the prism is equal to

$V=lwh$

Solve for w

That means-------> isolate the variable w

Divide both sides by (lh)

$V/(lh)=lwh/(lh)$

Simplify

$w=V/(lh)$

7 0

3 years ago

Help pleaseee guysss

Anvisha [2.4K]

Answer:

Step-by-step explanation:

f(x)=2x^2

f(x)=y

y=2x^2

y=2(-2)^2

y=2(4)

y=8

y=2x^2

y=2(-1)^2

Y=2

Y=2X^2

y=2(0)^2

y=0

y=2x^2

y=2(1)^2

y=2

y=2x^2

y=2(2)^2

y=8

DON'T FORGET BODMAS

6 0

3 years ago

There are 9 red, 5 yellow, and 7 green jellybeans in a jar. Without replacement what is the probability of choosing a green jell

Mrac [35]

Answer:

3/14 chance first draw

Step-by-step explanation:

There are 14 jellybeans in all, you won’t be replacing so the probability will change with each draw.

Probability the 1st draw is red = 3/14.

Probability the 2nd draw is red = (3–1)/(14–1) = 2/13.

Probability the 3rd draw is red = (2–1)/(13–1) = 1/12.

3/14 x 2/13 x 1/12 = 6/2184 = 1/364 is approximately 0.27%, pretty rare.

8 0

2 years ago

A theory predicts that the mean age of stars within a particular type of star cluster is 3.3 billion years, with a standard devi

Luden [163]

Answer:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

Step-by-step explanation:

Data given and notation

$\bar X=3.4$ represent the sample mean

$\sigma=0.4$ represent the population deviation for the sample

$n=50$ sample size

$\mu_o =3.3$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 3.3, the system of hypothesis would be:

Null hypothesis: $\mu \leq 3.3$

Alternative hypothesis: $\mu > 3.3$

Compute the test statistic

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$z=\frac{3.4-3.3}{\frac{0.4}{\sqrt{50}}}=1.768$

P value

Since is a one right tailed test the p value would be:

$p_v =P(z>1.768)=0.039$

6 0

2 years ago

In an all boys school, the heights of the student body are normally distributed with a mean of 70 inches and a standard deviatio

almond37 [142]

Answer:

0.010

Step-by-step explanation:

We solve the above question using z score formula

z = (x-μ)/σ, where

x is the raw score = 63 inches

μ is the population mean = 70 inches

σ is the population standard deviation = 3 inches

For x shorter than 63 inches = x < 63

Z score = x - μ/σ

= 63 - 70/3

= -2.33333

Probability value from Z-Table:

P(x<63) = 0.0098153

Approximately to the nearest thousandth = 0.010

Therefore, the probability that a randomly selected student will be shorter than 63 inches tall, to the nearest thousandth is 0.010.

3 0

2 years ago