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kondaur [170]
3 years ago
12

You are throwing darts at a dart board. You have a 16 chance of striking the bull's-eye each time you throw. If you throw 3 time

s, what is the probability that you will strike the bull's-eye all 3 times?
Mathematics
1 answer:
sveticcg [70]3 years ago
8 0
3 out of 48 i think its a rough estimate
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Use the Distributive Property to expand this expression.
viktelen [127]

Answer:

C is the correct answer

Step-by-step explanation:

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3 years ago
Read 2 more answers
Please help me I’m dying
Alex

Answer:

a° = 58°

b° = 48°

c = 74°

d° = 122°

Step-by-step explanation:

<em>Angles on a straight line add up to 180°</em>

58° + d° = 180°

d° = 180° - 58° = 122°

<em>Alternate angles are equal.</em>

a° = 58°

<em>Angles in a triangle add up to 180°.</em>

48° + 58° + c° = 180°

106° + c° = 180°

c° = 180° - 106° = 74°

<em>Angles on a straight line add up to 180°</em>

74° + 58° + b° = 180°

132° + b° = 180°

b° = 180° - 132° = 48°

5 0
3 years ago
The ice cream parlor serves spherical scoops of ice cream with a radius of 1 in. What is the approximate volume of 2 scoops of i
Varvara68 [4.7K]

Answer: 8.37 in³

Step-by-step explanation:

Given the following :

Shape of icre cream scoop = sphere

Radius (r) = 1 inch

Volume of a scoop of icecream = volume of sphere :

4/3πr³

(4/3) × 3.14 × 1³

1.333 × 3.14 × 1

= 4.1867 in³

Hence, the volume of 2 scoops of ice cream will be :

2 × ( volume of a scoop of ice cream)

(2 × 4.1867) in³

= 8.37 in³

6 0
3 years ago
What is the dominant emotional element in a cultural practice called?
Ber [7]

Answer:

Ideal Culture

8 0
3 years ago
A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls invol
bagirrra123 [75]

Answer:

a) 0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b) 0.118 = 11.8% probability that exactly 4 of the calls involve a fax message

c) 0.904 = 90.4% probability that at least 4 of the calls involve a fax message

d) 0.786 = 78.6% probability that more than 4 of the calls involve a fax message

Step-by-step explanation:

For each call, there are only two possible outcomes. Either it involves a fax message, or it does not. The probability of a call involving a fax message is independent of other calls. 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.

25% of the incoming calls involve fax messages

This means that p = 0.25

25 incoming calls.

This means that n = 25

a. What is the probability that at most 4 of the calls involve a fax message?

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4).

In which

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

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.001 + 0.006 + 0.025 + 0.064 + 0.118 = 0.214

0.214 = 21.4% probability that at most 4 of the calls involve a fax message

b. What is the probability that exactly 4 of the calls involve a fax message?

P(X = 4) = C_{25,4}.(0.25)^{4}.(0.75)^{21} = 0.118

0.118 = 11.8% probability that exactly 4 of the calls involve a fax message.

c. What is the probability that at least 4 of the calls involve a fax message?

Either less than 4 calls involve fax messages, or at least 4 do. The sum of the probabilities of these events is 1. So

P(X < 4) + P(X \geq 4) = 1

We want P(X \geq 4). Then

P(X \geq 4) = 1 - P(X < 4)

In which

P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

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

P(X = 0) = C_{25,0}.(0.25)^{0}.(0.75)^{25} = 0.001

P(X = 1) = C_{25,1}.(0.25)^{1}.(0.75)^{24} = 0.006

P(X = 2) = C_{25,2}.(0.25)^{2}.(0.75)^{23} = 0.025

P(X = 3) = C_{25,3}.(0.25)^{3}.(0.75)^{22} = 0.064

P(X

P(X \geq 4) = 1 - P(X < 4) = 1 - 0.096 = 0.904

0.904 = 90.4% probability that at least 4 of the calls involve a fax message.

d. What is the probability that more than 4 of the calls involve a fax message?

Very similar to c.

P(X \leq 4) + P(X > 4) = 1

From a), P(X \leq 4) = 0.214)

Then

P(X > 4) = 1 - 0.214 = 0.786

0.786 = 78.6% probability that more than 4 of the calls involve a fax message

8 0
3 years ago
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