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morpeh [17]
3 years ago
8

PLEASE I NEED HELP ASAP!!!

Mathematics
1 answer:
Rudiy273 years ago
6 0

No hill for one mile, uphill for one mile, downhill for one mile.

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Help help help help I have a lot of stuff to do and I need help
maw [93]

Answer:

2 5/8

Step-by-step explanation:

1 6/8 + 7/8 = 1 13/8 = 2 5/8.

6 0
2 years ago
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Can someone please help me with these!!!!!
Alona [7]
Yes, No, Yes, Yes, No. You fill in (x,y) numbers into the equations and if it cancels out into 4, it would be 4=4. If it doesn’t become 4=4, it’s no.
3 0
3 years ago
Whats the answer to 10×(2.5+13.5)
oee [108]

Simplify 2.5 + 13.5 to 16

10 × 16

Simplify

160

7 0
3 years ago
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The clearance between a pin and the collar around it is important for the proper performance of a disc drive for small computers
hichkok12 [17]

Answer:

(a) The distribution of (Y - X) is <em>N</em> (0.001, 0.0005).

(b) The probability that the pin will not fit inside the collar is 0.023.

Step-by-step explanation:

The random variable <em>X</em> is defined as the diameter of the pin and the random variable <em>Y</em> is defined as the diameter of the collar.

The distribution of <em>X</em> and <em>Y</em> is:

X\sim N(0.525, 0.0003)\\Y\sim N(0.526, 0.0004)

The random variables <em>X</em> and <em>Y</em> are independent of each other.

(a)

Compute the expected value of (Y - X) as follows:

E(Y-X)=E(Y)-E(X)=0.526-0.525=0.001

The mean of (Y - X) is 0.001.

Compute the variance of (Y - X) as follows:

V(Y-X)=V(Y)+V(X)-2Cov(X,Y)\\=V(Y)+V(X);\ X\ and\ Y\ are\ independent\\=0.0003^{2}+0.0004^{2}\\=0.00000025

SD(Y-X)=\sqrt{0.00000025}=0.0005

The standard deviation of (Y - X) is 0.0005.

Thus, the distribution of (Y - X) is <em>N</em> (0.001, 0.0005).

(b)

Compute the probability of [(Y - X) ≤ 0] as follows:

P(Y-X\leq 0)=P(\frac{(Y-X)-\mu_{Y-X}}{\sigma_{Y-X}}\leq \frac{0-0.001}{0.0005})=P(Z

*Use a <em>z</em>-table for the probability value.

Thus, the probability that the pin will not fit inside the collar is 0.023.

8 0
3 years ago
A cylindrical beaker with diameter 10 in and height 12 in. is filled with water that is then poured into a rectangular
emmasim [6.3K]

Answer:

A. The volume of the cylindrical beaker = 942.5 in3.

B. Volume of the pan = 378 in3

C. The water in the cylindrical beaker will over flow the pan.

D. The height of the water in the beaker after the pan is filled is 7.2 in

Step-by-step explanation:

Data given from the question.

Diameter = 10 in

Height = 12 in

Dimension of the pan = 14 in x 9 in x 3 in

Determination of the volume of each solid.

A. Volume of the cylindrical beaker:

Volume = πr2h

Radius (r) = 10/2 = 5 in

Height (h) = 12 in

Volume (V) =?

V = πr2h

V = π x (5)^2 x 12

V = 942.5 in3.

The volume of the cylindrical beaker = 942.5 in3.

B. Volume of the rectangular pan

Volume = Length x Width x Height

The dimension for the rectangular pan = volume of the pan = 14 in x 9 in x 3 in

Volume of the pan = 378 in3

C. Since the volume of the cylindrical beaker is higher than that of the pan, the water in the cylindrical beaker will overflow the pan.

D. Determination of the height of the water in the cylindrical beaker after the pan is filled. This is illustrated below.

Let us calculate the volume of the water in the cylindrical beaker after the pan is filled

The volume of the cylindrical beaker = 942.5 in3.

Volume of the pan = 378 in3

Volume of water in the cylinder beaker after the pan is filled = 942.5 - 378 = 564.5 in3

With this new volume, we can calculate the new height of the water as follow:

Volume = 564.5 in3

Radius (r) = 5 in (the radius remains the same)

Height (h) =?

V = πr2h

h = V/πr2

h = 564.5/π(5)^2

h = 7.2 in

Therefore, the height of the water in the beaker after the pan is filled is 7.2 in

4 0
2 years ago
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