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3 years ago

15

Consider the problem: A water bucket is being filled with water from a water faucet at a constant rate. When will the bucket be

full? What information would you need to be able to solve the problem?

1 answer:

chubhunter [2.5K]3 years ago

4 0

Answer:

To answer the problem you would need to know the rate in which water is going into the bucket (# quarts, gallons, whatever, and how fast) and how big the bucket is.

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7/8 - 5/12 ayudaaa por favor es para mañana

RUDIKE [14]

7/8 = 21/24
5/12 = 10/24

21/24-10/24 = 11/24

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3 years ago

Using the linear combination method what is the solution to the system of linear equations 7x - 2y = -20 and 9x + 4y = -6

Ghella [55]

x should equal -2 and y should equal -3 to solve this i multiplied the first equation by 2 then got rid of the 4y and -4y then i added the 14x to the 9 equaling 23x and added the -40 to the -6 which became 23x=-46 i solved this getting -s for x, then i plugged that into the equations to get y which was -3

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3 years ago

What is x in this equation −4.5= −0.5(x − 7.1)

Umnica [9.8K]

Answer:

x=16.1

Step-by-step explanation:

$-4.5= -0.5(x -7.1)$

Divide both sides by -0.5

$9=x-7.1$

Add 7.1 to both sides

$16.1=x$

Hope this helps!

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3 years ago

Estimate how long it will take to complete the

drek231 [11]

Answer:

14

Step-by-step explanation:

It's just that simple man.

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3 years ago

Samples of emissions from three suppliers are classified for conformance to air-quality specifications. The results from 100 sam

tigry1 [53]

Answer:

$P(A) = \frac{30}{100}$

$P(B) = \frac{77}{100}$

$P(A\ n\ B) = \frac{22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

Step-by-step explanation:

Given

See attachment for proper format of table

$n = 100$ --- Sample

A = Supplier 1

B = Conforms to specification

Solving (a): P(A)

Here, we only consider data in sample 1 row.

In this row:

$Yes = 22$ and $No = 8$

So, we have:

$n(A) = Yes + No$

$n(A) = 22 + 8$

$n(A) = 30$

P(A) is then calculated as:

$P(A) = \frac{n(A)}{Sample}$

$P(A) = \frac{30}{100}$

Solving (b): P(B)

Here, we only consider data in the Yes column.

In this column:

$(1) = 22$ $(2) = 25$ and $(3) = 30$

So, we have:

$n(B) = (1) + (2) + (3)$

$n(B) = 22 + 25 + 30$

$n(B) = 77$

P(B) is then calculated as:

$P(B) = \frac{n(B)}{Sample}$

$P(B) = \frac{77}{100}$

Solving (c): P(A n B)

Here, we only consider the similar cell in the yes column and sample 1 row.

This cell is: [Supplier 1][Yes]

And it is represented with; n(A n B)

So, we have:

$n(A\ n\ B) = 22$

The probability is then calculated as:

$P(A\ n\ B) = \frac{n(A\ n\ B)}{Sample}$

$P(A\ n\ B) = \frac{22}{100}$

Solving (d): P(A u B)

This is calculated as:

$P(A\ u\ B) = P(A) + P(B) - P(A\ n\ B)$

This gives:

$P(A\ u\ B) = \frac{30}{100} + \frac{77}{100} - \frac{22}{100}$

Take LCM

$P(A\ u\ B) = \frac{30+77-22}{100}$

$P(A\ u\ B) = \frac{85}{100}$

8 0

3 years ago