Fifty-nine thousandths<br> Write fifty-nine thousand in numerical format

2. In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviati

Kobotan [32]

Answer: $85,000

Step-by-step explanation:

Given : In a given population of two-earner male-female couples, male earnings have a mean of $40,000 per year and a standard deviation of $12,000.

$\mu_M=40,000\ \ ;\sigma_M=12,000$

Female earnings have a mean of $45,000 per year and a standard deviation of $18,000.

$\mu_F=45,000\ \ ;\sigma_F=18,000$

If C denote the combined earnings for a randomly selected couple.

Then, the mean of C will be :-

$\mu_c=\mu_M+\mu_F\\\\=40,000+45,000=85,000$

Hence, the mean of C = $85,000

8 0

3 years ago

In a standard deck of cards there 52 cards of which 26 are black and 26 are red. Additionally, there are 4 suits, hearts, clubs,

stich3 [128]

Using probability concepts, it is found that P(S and D) = 0.1275.

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A probability is the <u>number of desired outcomes divided by the number of desired outcomes</u>.
In a standard deck, there are 52 cards.
Of those, 13 are spades, and 13 are diamond.

The probability of selecting a spade with the first card is 13/52. Then, there is a 13/51 probability of selecting a diamond with the second. The same is valid for diamond then space, which means that the probability is multiplied by 2. Thus, the desired probability is:

$P(S \cap D) = 2 \times \frac{13}{52} \times \frac{13}{51} = \frac{2\times 13 \times 13}{52 \times 51} = 0.1275$

Thus, P(S and D) = 0.1275.

A similar problem is given at brainly.com/question/12873219

7 0

2 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.

6 0

3 years ago

-5/4-(-1/6) help please

tiny-mole [99]

A way to add fractions that always works is to multiply each numerator by the denominator of the other, then express the sum of products over the product of the denominators.
$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{d}{d}+\dfrac{c}{d}\cdot\dfrac{b}{b}\\\\=\dfrac{ad+bc}{bd}$

Here, you have
$\dfrac{-5}{4}-\dfrac{-1}{6}=\dfrac{-5\cdot6-(-1)\cdot4}{4\cdot6}=\dfrac{-26}{24}\\\\=\dfrac{-13}{12}=-1\frac{1}{12}$

The sum is -1 1/12

3 0

3 years ago

Hellllpme!!!!! ASAP!

Vikki [24]

Answer:

A

Step-by-step explanation:

This notation says that x+x+x+x+37=69

This can be simplified to say 4x+37=69

6 0

3 years ago

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Fifty-nine thousandths Write fifty-nine thousand in numerical format