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3 years ago

a lottery game has balls numbered 1 through 21. what is the probability of selecting a ball with a prime number on it or a 10

Mathematics

2 answers:

Ronch [10]3 years ago

6 0

Answer:

3/7

Step-by-step explanation:

The prime numbers that lie between 1 and 21 are 2,3,5,7,11,13,17, and 19. These 8 balls, plus the 10-ball comes to a total of 9 balls that could be picked out of the 21. Therefore, the probability of picking one of these balls is 9/21 or 3/7

Vinil7 [7]3 years ago

5 0

Answer:

$\frac{9}{21}$

Step-by-step explanation:

→ List the prime numbers between 1 and 21

2, 3, 5, 7, 11, 13, 17 and 19

→ Count how many ball that is

8 balls

→ Remember in the question it says "what is the probability of selecting a ball with a prime number on it or a 10"

8 balls + 1 ball (the one with the number 10)

→ 9 balls overall over a total of 21

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RideAnS [48]

Y=21/7

Explanation
7(y/7)=7(3/17)
y=7(3/17)
Multiple
y=21/7

4 0

3 years ago

Read 2 more answers

The cost of Matt and Natalie's dinner was $27.35. They want to leave a 20% tip. Which of the following is the closest to the amo

Ksenya-84 [330]

Answer:

Option D is correct.

Step-by-step explanation:

27.35 x 20/100

=> 2.735 x 2/1

=> $5.47

=> $5.50 (Rounded)

Therefore, Option D is correct.

Hoped this helped.

8 0

2 years ago

A survey of 90 men found that an average amount spent on St. Patrick's day of $55 with a standard deviation of $18. A similar su

dem82 [27]

Answer:

The value of the test statistic is 4.70.

Step-by-step explanation:

The hypothesis for this test can be defined as follows:

H₀: Men do not spend more than women on St. Patrick's day, i.e. μ₁ = μ₂.

Hₐ: Men spend more than women on St. Patrick's day, i.e. μ₁ > μ₂.

The population standard deviations are not known.

So a t-distribution will be used to perform the test.

The test statistic for the test of difference between mean is:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }$

Given:

$\bar x_{1}=55\\s_{1}=18\\n_{1}=90\\\bar x_{1}=44\\s_{1}=16\\n_{1}=86$

Compute the value of the test statistic as follows:

$t=\frac{\bar x_{1}-\bar x_{2}}{\sqrt{\frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}}} }=\frac{55-44}{\sqrt{\frac{15^{2}}{90}+\frac{16^{2}}{86} }}=4.70$

Thus, the value of the test statistic is 4.70.

7 0

3 years ago

Assuming that boy and girl babies are equally likely, what would be Kathy's probability of having at most one daughter if she

JulijaS [17]

Answer: 0.3125.

Step-by-step explanation:

The chance of having a boy is 0.5,
the chance of having a girl is also 0.5.

Outcome 1: all boys

boy boy boy boy

0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Outcome 2: first child is a girl, all others are boys

girl boy boy boy

0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Outcome 3: second child is a girl, all others are boys

boy girl boy boy

0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Outcome 4: third child is a girl, all others are boys

boy boy girl boy

0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Outcome 5: fourth child is a girl, all others are boys

boy boy boy girl

0.5 * 0.5 * 0.5 * 0.5 = 0.0625

Therefore,

0.0625 + 0.0625 + 0.0625 + 0.0625 + 0.0625 = 0.0625 * 5 = 0.3125

4 0

1 year ago

Factor 64v^3 + 192v^2 -56v - 168

Kaylis [27]

Hello from MrBillDoesMath!

Answer:

8(v+3) ( -1/2 (sqrt(14) - 4 v) (4 v + sqrt(14)) )

Discussion:

Given

64v^3 + 192v^2 - 56 v - 168

Factor 64v^2 from the first two terms. Factor 56 from the last two terms:

64v^2(v+3) - 56(v + 3) => factor (v+3) from both terms

(v+3) (64v^2 - 56) => factor 8 from both terms in the right ()

8(v+3)(8v^2-7) => factor 8y^2-7

8(v+3) ( -1/2 (sqrt(14) - 4 v) (4 v + sqrt(14)) )

Thank you,

MrB

4 0

3 years ago