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2 years ago

What goes in the question mark, greater than, less than, or equal? EASY STUFF

Mathematics

2 answers:

Ilya [14]2 years ago

8 0

Answer:

greater

Step-by-step explanation:

enot [183]2 years ago

3 0

Greater it’s rlly easy

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What are the intercepts of the line?

maxonik [38]

B.
x-intercept: –3; y-intercept: 7

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2 years ago

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Please help me thank you.

Alexxandr [17]

Answer:

Step-by-step explanation:

3 -4 | 2

0 -1 | 7

3/2 -2 | 1

R1/3 ----> R1 (divide Row 1 by 3)

1 -4/3 | 2/3

0 -1 | 7

3/2 -2 | 1

R3 - 3/2 R1 -------> R3 (multiply Row 1 by 3/2 and subtract it from Row 3)

1 -4/3 | 2/3

0 -1 | 7

0 0 | 0

R2 / -1 -------> R2 (Divide Row 2 by -1)

1 -4/3 | 2/3

0 1 | -7

0 0 | 0

R1 + 4/3 R2 --------> R1 (Multiply Row 2 by 4/3 and add it to Row 1)

1 0 | -26/3

0 1 | -7

0 0 | 0

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3 years ago

What is 625/1000 in the simplest form

Tatiana [17]

Your answer is 25/40 hope this helps ;)

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3 years ago

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The height of a right circular cylinder is 1.5 times the radius of the base. What is the ratio of the total surface area to the

Naily [24]

Let r represent the radius of cylinder.

We have been given that the height of a right circular cylinder is 1.5 times the radius of the base. So the height of the cylinder would be $1.5r$ .

We will use lateral surface area of pyramid to solve our given problem.

$LSA=2\pi r h$ , where,

LSA = Lateral surface area of pyramid,

r = Radius,

h = height.

Upon substituting our given values in above formula, we will get:

$LSA=2\pi r\cdot (1.5)r$

Now we will find the total surface area of cylinder.

$TSA=2\pi r(r+h)$

$TSA=2\pi r(r+1.5r)$

$TSA=2\pi r(2.5r)$

$\frac{TSA}{LSA}=\frac{2\pi r(2.5r)}{2\pi r(1.5r)}$

$\frac{TSA}{LSA}=\frac{2.5r}{1.5r}$

$\frac{TSA}{LSA}=\frac{25}{15}$

$\frac{TSA}{LSA}=\frac{5}{3}$

Therefore, the ratio of total surface area to lateral surface area is $5:3$ .

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2 years ago

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil

Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

$H_0$ : The true average penetration is 50 mils

$H_a$ : The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the level of confidence is α = 0.05 then the $z_\alpha$ score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)