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2 years ago

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Timothy’s mom bought 7/8 of a pound of grapes. She wants to split it between 8 kids. What fraction of the grapes will each kid g

et?
A. 1/8

B. 7/64

C. 9/64

2 answers:

I am Lyosha [343]2 years ago

8 0

Answer:

Option B

7/64

Step-by-step explanation:

Since the she bought 7/8 of a pound of grapes, the amount of grapes that Timothy's mum will be sharing is only 7/8 of a pound

Total number of kids = 6

To get the amount that each kid will get, we will have to divide 7/8 by 6

Hence we have

$\frac{7}{8} \times \frac{1}{6}= \frac{7}{64}$

Hence, each child will get 7/64 of a pound of grapes, if the grapes are shared equally.

Anni [7]2 years ago

6 0

Answer:

B. 7/64

Step-by-step explanation:

Since Timothy's mom bought 7/8 of a pound of grapes, the amount of grapes that she will split between 8 kids will be:

7/8 ÷ 1/8 = 7/64

Therefore, each kid will get 7/64 of a pound of grapes, if the grapes are splitted equally.

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I don't understand the question you are asking sorry.

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2 years ago

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol

bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

25% of the incoming calls involve fax messages

This means that $p = 0.25$

25 incoming calls.

This means that $n = 25$

a. What is the probability that at most 4 of the calls involve a fax message?

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

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214$

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

$P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118$

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

$P(X < 4) + P(X \geq 4) = 1$

We want $P(X \geq 4)$ . Then

$P(X \geq 4) = 1 - P(X < 4)$

In which

$P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001$

$P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006$

$P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025$

$P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064$

$P(X$

$P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904$

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

$P(X \leq 4) + P(X > 4) = 1$

From a), $P(X \leq 4) = 0.214)$

Then

$P(X > 4) = 1 - 0.214 = 0.786$

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

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2 years ago

Which of the following predictions can be calculated using a geometric distribution?

snow_lady [41]

Answer: C

both a and b

Step-by-step explanation:

Both options A and B deals with the number of trials required for a single success. Thus, they are negative binomial distribution where the number of successes (r) is equal to 1.

The geometric distribution is a special case of the negative binomial distribution that deals with the number of trials required for a single success.

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2 years ago

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A bicycle wheel with diameter 16 inches rides over a screw in the street. The screw is on level ground before it punctures the b

Julli [10]

Answer:

The correct option is;

16 inches

Step-by-step explanation:

The parameters of the motion given are;

The diameter, D of the bicycle = 16 inches;

The distance the bike moves (forward) after the screw punctures the tire = 56·π inches

We note that the circumference of the bicycle = π·D = π × 16 = 16·π inches

Therefore;

56·π inches/(16·π inches) = 3.5

Showing that the bicycle moves three and half complete turns (revolution) where after each complete turn, the screw starts from the bottom of the tire.

The height, h of the screw in the final half turn is given by the relation;

h = A×cos(Bx - C) + D

A = Amplitude of the motion = Diameter/2 = 16/2 = 8

P = The period of the motion 2·π/B

B·x = The angle described by the motion = Half of one revolution = π = 180°

C = Phase shift = π

D = The midline = Diameter/2 = 8 inches

Therefore;

h = 8×cos(π - π) + 8 = 16 inches

After the bike moves forward another 56·π inches the height of the screw = 16 inches.

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2 years ago

What is the fifth term of the geometric sequence?<br> a1 = 120, a2 = 36, a3 = 10.8, a6 = 0.2916

Alisiya [41]

R = a2/a1 = 36/120 = 0.2

a5 = a1 * r^4 = 120 * 0.2^4 = 120 * 0.0016 = 0.192

Hope it helps!

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2 years ago

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