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Maksim231197 [3]

2 years ago

5

A box contains 2 plain pencils and 2 pens. A second box contains 1 color pencil and 3 crayons. One item from each box is chosen

at random. What is the probability that a pen from the first box and a crayon from the second box are selected?

1 answer:

Nitella [24]2 years ago

6 0

I know the answer but I dont know how to describe it, sorry

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I need help with this question.

scZoUnD [109]

Answer:

8

Step-by-step explanation:

don't ask how I got it I looked it up

4 0

2 years ago

Rajan needs 1 liter of 15% hydrochloric acid solution for an experiment. He checks his storage cabinet, but he only has a 5-lite

yarga [219]

Answer:

<em>Rajan must use </em><em>0.75</em><em> liters of 5% hydrochloric acid solution and </em><em>0.25</em><em> liters of 45% hydrochloric acid solution.</em>

Step-by-step explanation:

Let us assume that, x liters of the 5% hydrochloric acid and y liters of the 45% hydrochloric acid solutions are combined.

As Rajan need total of 1 liter of solution, so

$x+y=1$

i.e $x=1-y$ --------------------1

As Rajan needs 5% hydrochloric acid and 45% hydrochloric acid to make a 1 liter batch of 15% hydrochloric acid, hence acid content of the mixture of two acids will be same as of the final one, so

$0.05x+0.45y=1\times 0.15$

i.e $5x+45y=15$ -------------2

Putting value of x from equation 1 in equation 2,

$\Rightarrow 5(1-y)+45y=15$

$\Rightarrow 5-5y+45y=15$

$\Rightarrow 5+40y=15$

$\Rightarrow 40y=15-5=10$

$\Rightarrow y=0.25$

Putting the value of y in equation 1,

$x=1-0.25=0.75$

Therefore, Rajan must use 0.75 liters of 5% hydrochloric acid solution and 0.25 liters of 45% hydrochloric acid solution.

5 0

3 years ago

ALI. calonades i TONS<br>QUESTIONS<br>1 begun bow PORT<br>RT pod is<br>8.1.1Give a reason why P2=70

kolbaska11 [484]

Can you put a picture of the question so that i can help you out

6 0

3 years ago

Round the number 206.8 to the underline place the 0 is underlined

lara [203]

Answer:

210.0

Step-by-step explanation:

you are rounding to the tens place so you look at the number before it is greater than 5 so you round up getting your answer of 210.0

8 0

2 years ago

Read 2 more answers

Calls to a customer service center last on average 2.8 minutes with a standard deviation of 1.4 minutes. An operator in the call

Natali5045456 [20]

Answer:

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n-values of normal variable:

Suppose we have n values from a normally distributed variable. The mean of the sum of all the instances is $M = n\mu$ and the standard deviation is $s = \sigma\sqrt{n}$

Calls to a customer service center last on average 2.8 minutes.

This means that $\mu = 2.8$

75 calls each day.

This means that $n = 75$

What is the expected total amount of time in minutes the operator will spend on the calls each day

This is M, so:

$M = n\mu = 75*2.8 = 210$

The expected total amount of time the operator will spend on the calls each day is of 210 minutes.

6 0

3 years ago