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2 years ago

PLEASE HELP ME!!! GIVING BRAINLIEST

Mathematics

1 answer:

KIM [24]2 years ago

4 0

Hi there! I am here to help with your question.

I believe the answer is A, 39°, 57 minutes 23.76 seconds.

Thank you and I hope this helped.

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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

What is the ansser <br><br><br> plase help

aniked [119]

Answer:

sir this isnt math

Step-by-step explanation:

put it in spanish next time

5 0

2 years ago

Read 2 more answers

Sketch the graph for each function. Choose either A, B, C, or D.

Serjik [45]

Hi. Here it is!
I hope this help...

—> And the answer is letter A

6 0

2 years ago

Adult men have heights that a normally distributed with a mean of 69.5 inches and a standard deviation of 2.4 inches. Adult wome

Aliun [14]

Answer:

Step-by-step explanation:

From the information given:

For Adult Men

Mean $\mu$ = 69.5

Standard deviation $\sigma$ = 2.4

observed value X = 74

For Adult Women

Mean $\mu$ = 63.8

Standard deviation $\sigma$ = 2.6

observed value X = 70

Therefore ; the values for their z scores can be obtained in order to determine who is more unusually tall within his or her respective sex

For Adult Men :

$z = \dfrac{X- \mu}{\sigma}$

$z = \dfrac{74- 69.5}{2.4}$

$z = \dfrac{4.5}{2.4}$

z = 1.875

For Adult Women :

$z = \dfrac{X- \mu}{\sigma}$

$z = \dfrac{70- 63.8}{2.6}$

$z = \dfrac{6.2}{2.6}$

z = 2.3846

Thus; we can conclude that , the women is more unusually tall within his or her respective sex

5 0

3 years ago

We are going to sell Harlem Sweatshirts to raise money for

son4ous [18]

Answer:

32 sweatshirts

Step-by-step explanation:

profit per sweatshirt 28-13=$15

475/15=31.66667

3 0

3 years ago