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erik [133]
2 years ago
10

In a shipment of 850 widgets, 32 are found to be defective. At this rate, how many defective widgets could be expected in 22,000

widgets? Round to the nearest whole number.
Mathematics
1 answer:
Kobotan [32]2 years ago
3 0
You can solve this using fractions, and say that 32/850 are defective. To find the amount of defective widgets out of 22,000, we can then multiply to make the denominator 22,000. To work out what we need to multiply by, we can just do 22,000 ÷ 850 = 25.88. Now if we multiply 32 by 25.88, we get our answer as 828 widgets.

I hope this helps!
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At the time of her grandson's birth, a grandmother deposits $12,000.00 in an account that pays 2% compound monthly. What will be that value of the account at the child's twenty-first birthday, assuming that no other deposits or withdrawls are made during the period.
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A(t) = P(1+(r/n))^(nt)
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kenny6666 [7]

Answer:

a: 4x+4x, 8x

b:6+24y,6*1+6*4y

Step-by-step explanation:

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2 years ago
What is the value of 5 in 756? Write and draw to explain how you know
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The 5 is represented as 50. The number is in the tens place, and 5 in the tens place means 50.
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6.8 Use the Normal approximation. Suppose we toss a fair coin 100 times. Use the Normal approximation to find the probability th
Maru [420]

Answer:

(a) The probability that proportion of heads is between 0.30 and 0.70 is 1.

(b) The probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

Step-by-step explanation:

Let <em>X</em> = number of heads.

The probability that a head occurs in a toss of a coin is, <em>p</em> = 0.50.

The coin was tossed <em>n</em> = 100 times.

A random toss's result is independent of the other tosses.

The random variable <em>X</em> follows a Binomial distribution with parameters n = 100 and <em>p</em> = 0.50.

But the sample selected is too large and the probability of success is 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of \hat p<em> </em>(sample proportion of <em>X</em>) if the following conditions are satisfied:

  1. np ≥ 10
  2. n(1 - p) ≥ 10

Check the conditions as follows:

 np=100\times 0.50=50>10\\n(1-p)=100\times (1-0.50)=50>10

Thus, a Normal approximation to binomial can be applied.

So,  \hat p\sim N(p,\ \frac{p(1-p)}{n})

\mu_{p}=p=0.50\\\sigma_{p}=\sqrt{\frac{p(1-p)}{n}}=0.05

(a)

Compute the probability that proportion of heads is between 0.30 and 0.70 as follows:

P(0.30

                              =P(-4

Thus, the probability that proportion of heads is between 0.30 and 0.70 is 1.

(b)

Compute the probability that proportion of heads is between 0.40 and 0.65 as follows:

P(0.40

                              =P(-2

Thus, the probability that proportion of heads is between 0.40 and 0.65 is 0.9759.

6 0
2 years ago
two friends went rowing on the river. they traveled with current for 3 hours. the way back took them 4 hours and 48 minutes. fin
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Answer:

1.5 mph

Step-by-step explanation:

The speed in the still water = 6.5 mph.

The speed of the current = x mph.

1. Two friends traveled with current for 3 hours. The speed with the current is 6.5 + x mph, so they traveled 3(x+6.5) miles.

2. Two friends traveled against the current for 4 hours and 48 minutes that is 4\dfrac{4}{5}=4.8 hours. The speed against the current is 6.5 - x mph, so they traveled 4.8(6.5-x) miles.

They traveled the same distance with the current and against the current, hence,

3(x+6.5)=4.8(6.5-x)

Solve this equation:

3x+19.5=31.2-4.8x\\ \\3x+4.8x=31.2-19.5\\ \\7.8x=11.7\\ \\x=1.5\ mph

4 0
2 years ago
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