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3 years ago

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9. Eric guessed that Troop A was going to have 23 members in 2011 when they actually had 20. How much was Eric off by in his est

imate?
a.20%b. 87%c. 13% d. 15%

1 answer:

julia-pushkina [17]3 years ago

8 0

Answer:The answer is D

Step-by-step explanation:

23 - 20=3

(3 ➗ 20) *100

= 3 ✖ 5

=15%

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A researcher can hypothesize that the first mean is larger than the second (H1: μ1 > μ2 ), that the first mean is smaller than the second (H1: μ1 < μ2 ), or that the means are different (H1: μ1 ≠ μ2 ). These ae the alternative hypothesis.

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3 years ago

The probability that a student has a Visa card (event V) is .63. The probability that a student has a MasterCard (event M) is .1

Aleksandr-060686 [28]

Answer:

a) The probability that a student has either a Visa card or a MasterCard is 0.71.

b) V and M are not independent.

Step-by-step explanation:

Given : The probability that a student has a Visa card (event V) is 0.63. The probability that a student has a MasterCard (event M) is 0.11. The probability that a student has both cards is 0.03.

To find :

a) The probability that a student has either a Visa card or a MasterCard ?

b) In this problem, are V and M independent ?

Solution :

The probability that a student has a visa card(event V) is P(V)= 0.63

The probability that a student has a MasterCard (event M) is P(M)= 0.11

The probability that a student has both cards is $P(V \cap M)=0.03$

a) Probability that a student has either a Visa card or a Master Card is given by,

$P(V \cup M) = P(V) + P(M) - P(V\cap M)$

$P(V \cup M) = 0.63+ 0.11- 0.03$

$P(V \cup M) =0.74- 0.03$

$P(V \cup M) =0.71$

The probability that a student has either a Visa card or a MasterCard is 0.71.

b) Two events, A and B, are independent if $P(A\cap B)=P(A)P(B)$

For V and M to be independent the condition is satisfied,

$P(V\cap M)=P(V)P(M)$

Substitute the values,

$0.03=0.63\times 0.11$

$0.03\neq 0.0693$

So, V and M are not independent.

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3 years ago