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3 years ago

Andrew must cut a rope 9 1/7 yards long into 8 equal

Mathematics

2 answers:

Fittoniya [83]3 years ago

6 0

Answer:

each rope should be 1 1/7 yard long

Step-by-step explanation:

9 1/7 ÷8

1 1/7

Vesna [10]3 years ago

3 0

Answer:

8/7 yd per piece

Step-by-step explanation:

To answer this, divide the total rope length 9 1/7 yd by 8 pieces:

64 yd

------------------ = 8/7 yd per piece

7(8 pieces)

Check: does 8 times (8/7 yd/piece) come out to 64/7 yd, or 9 1/7 yd? YES

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The charity gala costs $5.17 for every attendee. How many attendees can there be, at most, if the budget for the charity gala is

Elodia [21]

Answer:

15 people are able to be at the charity gala because 77:55 divided by 5:17.

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3 years ago

How many 4-digit numbers divisible by 5, all of the digits of which are odd, are there

HACTEHA [7]

Answer:

So we must create 4-digit numbers with the 5 odd digits in the image.

First, a number is only divisible by 5 if the last digit is 0 or 5, and we only can use the last digit equal to 5 (because 0 is not in the image).

if each odd number can be used only once, we have:

Now, for the other 3 digits we have 4 options.

So for the first one we have 4 options,

for the second we have 3 options (because one is already taken)

for the last one we have only 2 options

then the number of combinations is:

C = 4*3*2 = 24.

If the numbers can be repeated (for example, 5555 is allowed, then)

we still have our last digit fixed in 5, and for the first digit we have 5 options, for the second we also have 5 options, and for the third we also have 5 options, then we have a total of:

C = 5*5*5 = 25*5 = 125 combinations.

3 0

4 years ago

The sum of 21, 1.7,0.25

Scorpion4ik [409]

21+1.7+0.25=22.95

You add 21+1.7, which equals 22.7, then add 0.25 to get 22.95.

8 0

3 years ago

5.53 An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day. What

Licemer1 [7]

Answer:

a) 38.4% probability that on a given day this item is requested more than 5 times.

b) 0.67% probability that on a given day this item is not requested at all.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

An inventory study determines that, on average, demands for a particular item at a warehouse are made 5 times per day.

This means that $\mu = 5$

What is the probability that on a given day this item is requested

(a) more than 5 times?

Either it is requested 5 times or less, or it is requested more than 5 times. The sum of the probabilities of these events is decimal 1. So

$P(X \leq 5) + P(X > 5) = 1$

We want P(X > 5). So

$P(X > 5) = 1 - P(X \leq 5)$

In which

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

$P(X = 1) = \frac{e^{-5}*(5)^{1}}{(1)!} = 0.0337$

$P(X = 2) = \frac{e^{-5}*(5)^{2}}{(2)!} = 0.0842$

$P(X = 3) = \frac{e^{-5}*(5)^{3}}{(3)!} = 0.1404$

$P(X = 4) = \frac{e^{-5}*(5)^{4}}{(4)!} = 0.1755$

$P(X = 5) = \frac{e^{-5}*(5)^{5}}{(5)!} = 0.1755$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.0067 + 0.0337 + 0.0842 + 0.1404 + 0.1755 + 0.1755 = 0.616$

$P(X > 5) = 1 - P(X \leq 5) = 1 - 0.616 = 0.384$

38.4% probability that on a given day this item is requested more than 5 times.

(b) not at all?

$P(X = 0) = \frac{e^{-5}*(5)^{0}}{(0)!} = 0.0067$

0.67% probability that on a given day this item is not requested at all.

3 0

4 years ago

What is the exact area of a circle with a diameter of 25

antiseptic1488 [7]

$\pi (25/2) ^{2}$ =490.873

7 0

4 years ago