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What quantity of a 30% solution do you need to make 150 mL of a 20% solution?

Bumek [7]

Answer:

Volume required = 100mL

Step-by-step explanation:

Let the volume of 30% solution required be = V mL

Amount of solute in V mL of solution = 30% of V = 0.3V...............(i)

This amount is also present in 150 mL of 20% solution

For 150 mL of 20% solution we have amount of solute = 150 mL X 20%

= 30..............(ii)

Thus equating i and ii we get

0.3V = 30

Thus V =100 mL

7 0

3 years ago

Joe is a repair technician for a computer company. Each week he has a batch of computers that needs fixed. The number of compute

mrs_skeptik [129]

Answer:

D

Step-by-step explanation:

4 0

3 years ago

G(b) = 5b- 9 h(b) = (b - 1)^2 Evaluate. (hog)(-6) =

laiz [17]

Given:

The two functions are:

$g(b)=5b-9$

$h(b)=(b-1)^2$

To find:

The value of $(h\circ g)(-6)$ .

Solution:

We have,

$g(b)=5b-9$

$h(b)=(b-1)^2$

We know that,

$(h\circ g)(b)=h(g(b))$

$(h\circ g)(b)=h(5b-9)$

$(h\circ g)(b)=[(5b-9)-1]^2$

$(h\circ g)(b)=[5b-10]^2$

Putting $b=-6$ , we get

$(h\circ g)(-6)=[5(-6)-10]^2$

$(h\circ g)(-6)=[-39-10]^2$

$(h\circ g)(-6)=[-49]^2$

$(h\circ g)(-6)=2401$

Therefore, the value of $(h\circ g)(-6)$ is 2401.

7 0

3 years ago

: A personal computer manufacturer buys 36% of its chips from Japan and the rest from the United St...

meriva

Answer: 0.9862

Step-by-step explanation:

Given : The probability that the chips belongs to Japan: P(J)= 0.36

The probability that the chips belongs to United States : P(U)= 1-0.36=0.64

The proportion of Japanese chips are defective : P(D|J)=0.017

The proportion of American chips are defective : P(D|U)=0.012

Using law of total probability , we have

$P(D)=P(D|J)\times P(J)+P(D|U)\times P(U)\\\\\Rightarrow\ P(D)=0.017\times0.36+0.012\times0.64=0.0138$

Thus , the probability that chip is defective = 0.0138

Then , the probability that a chip is defect-free= $1-0.0138=0.9862$

3 0

3 years ago

Consider a single spin on the spinner shown below. A circle is split into sections 1, 2, 4, and 3. A spinner is pointing at numb

vladimir2022 [97]

Answer:

The answer is "Options A, B, and E represent mutually exclusive events".

Step-by-step explanation:

Two occurrences that can happen immediately called mutually incompatible. Let's now glance at our options and figure out where the statements are mutually incompatible events.

In Option A: You could see that landing on an unwanted portion and arriving on 2 are events that are locally incompatible, even as undesirable portion contains 3 and 4, and 2 were shaded.

In Option B: Arriving on a shaded part and falling on 3 are also mutually incompatible because there are 3 on a windows azure.

In Option C: A darkened portion and an increasing amount can land while 2 would be an even number as well as on the shaded portion. That number is very much the same.

In Option D: At the same time as 4 is greater than 3 and it is situated upon an undistressed section, landing and attracting a number larger than 3 can happen.

In Option E: Landing on a shaded part and landing on even a shaded part is an excluding event, since shaders may either be shaded or unlit.

3 0

3 years ago

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