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2 years ago

The cube has a volume of 64 cubic inches. What is the length of a side of the cube? Explain. PLEASE HURRY!

Mathematics

2 answers:

Mashcka [7]2 years ago

6 0

4

A^3 = 64

Therefore a = 4

Nostrana [21]2 years ago

5 0

Answer:

Step-by-step explanation:

a^3 = 64

therefore , a = 4

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A one-celled organism measures 32 mm in length in a photograph. If the photo has been enlarged by a factor of 100, what is the a

larisa86 [58]

32 mm = length × 100

(32 mm)/100 = length . . . . . divide both sides by 100

0.32 mm = length

The actual length of the organism is 0.32 mm.

5 0

2 years ago

Read 2 more answers

Solve for x, y, and z

zloy xaker [14]

Answer:

The solution is x=1,y=2,z=3

Step-by-step explanation:

The given system of equations is ;\

2x−3y+4z=8...(1)

3x+4y−5z=−4...(2)

4x−5y+6z=12...(3)

Make x the subject in equation (1)

$x=\frac{8+3y-4z}{2}...(4)$

Put equation (4) into equation (2) and (3)

$3(\frac{8+3y-4z}{2})+4y-5z=-4$

Multiply through by;

$3(8+3y-4z)+8y-10z=-8$

Expand;

$24+9y-12z+8y-10z=-8$

Simplify;

$17y-22z=-32...(5)$

Equation (4) in (3)

$4(\frac{8+3y-4z}{2})-5y+6z=12$

$2(8+3y-4z)-5y+6z=12$

$16+6y-8z-5y+6z=12$

$y-2z=-4$

$y=2z-4...(6)$

Put equation (6) into equation (5)

$17(2z-4)-22z=-32$

$34z-68-22z=-32$

$34z-22z=-32+68$

$12z=36$

z=3

Put z=3 into equation (6)

y=2(3)-4=2

Put y=2 and z=3 into equation 4

$x=\frac{8+3(2)-4(3)}{2}=1$

The solution is x=1,y=2,z=3

3 0

2 years ago

A research study investigated differences between male and female students. Based on the study results, we can assume the popula

oksian1 [2.3K]

Answer: 0.0548

Step-by-step explanation:

Given, A research study investigated differences between male and female students. Based on the study results, we can assume the population mean and standard deviation for the GPA of male students are µ = 3.5 and σ = 0.05.

Let $\overline{X}$ represents the sample mean GPA for each student.

Then, the probability that the random sample of 100 male students has a mean GPA greater than 3.42:

$P(\overline{X}>3.42)=P(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}>\dfrac{3.42-3.5}{\dfrac{0.5}{\sqrt{100}}})\\\\=P(Z>\dfrac{-0.08}{\dfrac{0.5}{10}})\ \ \ [Z=\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(Z>1.6)\\\\=1-P(Z$