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3 years ago

Help please best answer

Mathematics

1 answer:

Anni [7]3 years ago

5 0

Hey!

Hope this helps...

~~~~~~~~~~~~~~~~~~~~~~~~~

When looking at this graph we see 2 things:

(1) How they made the graph, looks awful, anyway...

(2) They practically give you the answer...

When finding an answer to a question like this, you want to use a formula (like this):

A = t * p (where A is the amount of people, t is total amount of people, and p is percent of people / 100)...

So... we fill in the givens...

We know t = 400, and (by looking at the graph) we see that p is around 0.10 and 0.15. So for our sake we'll make p = 0.13 (about halfway)...

A = 400 * 0.13

A ~ 52

The answer is: About 52 people bought 20 items...

And because we got 52, and 52 is close to 54...

The Answer: B.) 54

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mean=134.3

median=65.5

mode=60

Step-by-step explanation:

Mean is computing by adding the all data values and then divided by number of values

mean=sum of all values/number of values

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41 45 45 50 50 55 60 60 60 65 66 71 88 110 125 235 250 350 360 500

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4 years ago

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Probability that fewer than 2 of these parts are defective is 0.604.

Step-by-step explanation:

We are given that at a certain auto parts manufacturer, the Quality Control division has determined that one of the machines produces defective parts 19% of the time.

A random sample of 7 parts produced by this machine is chosen.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 7 parts

r = number of success = fewer than 2

p = probability of success which in our question is % of defective

parts produced by one of the machine, i.e; 19%

LET X = Number of parts that are defective

So, it means X ~ Binom(n = 7, p = 0.19)

Now, probability that fewer than 2 of these parts are defective is given by = P(X < 2)

P(X < 2) = P(X = 0) + P(X = 1)

= $\binom{7}{0}\times 0.19^{0} \times (1-0.19)^{7-0}+ \binom{7}{1}\times 0.19^{1} \times (1-0.19)^{7-1}$

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