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Alexus [3.1K]
3 years ago
14

10 2/5 divided by 7 please help

Mathematics
1 answer:
AveGali [126]3 years ago
3 0
1.48 because 10 and 2/5 is 52/5 is 10.4 from there divide by 7 which is 1.48
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Determine the measure of arc CED. <br><br> A. 180°<br> B. 99°<br> C. 198°<br> D. 81°
Lina20 [59]

Answer:

C

Step-by-step explanation:

An angle whose vertex lies on a circle and whose sides are 2 chords of the circle is one half the measure of the intercepted arc.

arc CED is the intercepted arc, hence

arc CED = 2 × 99° = 198°

3 0
3 years ago
Suppose that 50% of all young adults prefer McDonald's to Burger King when asked to state a preference. A group of 12 young adul
ddd [48]

Answer:

a) 0.194 = 19.4% probability that more than 7 preferred McDonald's

b) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

c) 0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

Step-by-step explanation:

For each young adult, there are only two possible outcomes. Either they prefer McDonalds, or they prefer burger king. The probability of an adult prefering McDonalds is independent from other adults. 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.

50% of all young adults prefer McDonald's to Burger King when asked to state a preference.

This means that p = 0.5

12 young adults were randomly selected

This means that n = 12

(a) What is the probability that more than 7 preferred McDonald's?

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12)

In which

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

P(X = 8) = C_{12,8}.(0.5)^{8}.(0.5)^{4} = 0.121

P(X = 9) = C_{12,9}.(0.5)^{9}.(0.5)^{3} = 0.054

P(X = 10) = C_{12,10}.(0.5)^{10}.(0.5)^{2} = 0.016

P(X = 11) = C_{12,11}.(0.5)^{11}.(0.5)^{1} = 0.003

P(X = 12) = C_{12,12}.(0.5)^{12}.(0.5)^{0} = 0.000

P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) = 0.121 + 0.054 + 0.016 + 0.003 + 0.000 = 0.194

0.194 = 19.4% probability that more than 7 preferred McDonald's

(b) What is the probability that between 3 and 7 (inclusive) preferred McDonald's?

P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

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

P(X = 3) = C_{12,3}.(0.5)^{3}.(0.5)^{9} = 0.054

P(X = 4) = C_{12,4}.(0.5)^{4}.(0.5)^{8} = 0.121

P(X = 5) = C_{12,5}.(0.5)^{5}.(0.5)^{7} = 0.193

P(X = 6) = C_{12,6}.(0.5)^{6}.(0.5)^{6} = 0.226

P(X = 7) = C_{12,7}.(0.5)^{7}.(0.5)^{5} = 0.193

P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.054 + 0.121 + 0.193 + 0.226 + 0.193 = 0.787

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred McDonald's

(c) What is the probability that between 3 and 7 (inclusive) preferred Burger King?

Since p = 1-p = 0.5, this is the same as b) above.

So

0.787 = 78.7% probability that between 3 and 7 (inclusive) preferred Burger King

7 0
3 years ago
Name the property that the following statement illustrates.
vodomira [7]

Answer:

(D.) Commutative Property of Addition.

Step-by-step explanation:

The rule of Commutative Property of Addition is as follows:

a + b = b + a,

so in theory, if the case were 2 + 3 = 3 + 2, they would be the same.

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Assoli18 [71]

Answer:

A. As the Study time in increases the Test score increases.

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3 years ago
Ii) p – (p – q) – q – (q – p)
sashaice [31]

Answer:

-q + p

Step-by-step explanation:

p - (p - q) - q - (q - p) = p - p + <u>q -q</u> - q + p

                                = 0 + 0 - q + p

                               = -q + p

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4 years ago
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