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3 years ago

Byron has 72 coins in his piggy bank. The piggy bank contains only dimes

Mathematics

2 answers:

sertanlavr [38]3 years ago

4 0

Answer:

(3) $0.25q+0.10(72-q)=14.70$

Step-by-step explanation:

Let d represent number of dimes and q represent number of quarters.

We have been given that Byron has 72 coins in his piggy bank. We can represent this information in an equation as:

$q+d=72...(1)$

We are also told that the value of all coins is $14.70.

Value of d dimes: $0.10d$ .

Value of q quarters: $0.25q$ .

$0.25q+0.10d=14.70...(2)$

From equation (1), we will get:

$d=72-q$

Substitute this value in equation (2):

$0.25q+0.10(72-q)=14.70$

Therefore, the equation $0.25q+0.10(72-q)=14.70$ can be used to determine q, the number of quarters.

Mashutka [201]3 years ago

3 0

3. 25 cents per quarter, and then the reamaining amount ids dimes, which are 10 cents apeice.

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WILL GOVE BRAINIEST Kim and Tegan train for the iron man on the weekend. They bike at a constant rate of miles per hour. The tab

Rus_ich [418]

Answer:

C 210

Step-by-step explanation:

To solve this, First we must take 42/3, Since it equals 14, we take 15x14 and find 210

(edit: to solve for Kim's missing values, you do the same with 2 and 32)

6 0

2 years ago

After a large scale earthquake, it is predicted that 15% of all buildings have been structurally compromised.a) What is the prob

Westkost [7]

Answer:

a) 13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) 17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) 17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) 75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

Step-by-step explanation:

For each building, there are only two possible outcomes after a earthquake. Either they have been damaged, or they have not. The probability of a building being damaged is independent from other buildings. 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.

15% of all buildings have been structurally compromised.

This means that $p = 0.15$

20 buildings

This means that $n = 20$

a) What is the probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised?

This is P(X = 1).

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

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

13.68% probability that if engineers inspect 20 buildings they will find exactly one that is structurally compromised.

b) What is the probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised?

$P(X < 2) = P(X = 0) + P(X = 1)$

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

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0388 + 0.1368 = 0.1756$

17.56% probability that if engineers inspect 20 buildings they will find less than 2 that are structurally compromised

c) What is the probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised?

Either they find 4 or less, or they find more than 4. The sum of the probabilities of these events is 1. So

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

$P(X > 4) = 1 - P(X \leq 4)$

In which

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

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

$P(X = 0) = C_{20,0}.(0.15)^{0}.(0.85)^{20} = 0.0388$

$P(X = 1) = C_{20,1}.(0.15)^{1}.(0.85)^{19} = 0.1368$

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.0388 + 0.1368 + 02293 + 0.2428 + 0.1821 = 0.8298$

$P(X > 4) = 1 - P(X \leq 4) = 1 - 0.8298 = 0.1702$

17.02% probability that if engineers inspect 20 buildings they will find greater than 4 that are structurally compromised

d) What is the probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised?

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

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

$P(X = 2) = C_{20,2}.(0.15)^{2}.(0.85)^{18} = 0.2293$

$P(X = 3) = C_{20,3}.(0.15)^{3}.(0.85)^{17} = 0.2428$

$P(X = 4) = C_{20,4}.(0.15)^{4}.(0.85)^{16} = 0.1821$

$P(X = 5) = C_{20,5}.(0.15)^{5}.(0.85)^{15} = 0.1028$

$P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.2293 + 0.2428 + 0.1821 + 0.1028 = 0.7570$

75.70% probability that if engineers inspect 20 buildings they will find between 2 and 5 (inclusive) that are structurally compromised

e) What is the expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings?

The expected value of the binomial distribution is:

$E(X) = np$

$E(X) = 20*0.15 = 3$

The expected number of buildings that an engineer will find structurally compromised if the engineer inspects 20 buildings is 3.

3 0

3 years ago

NEED HELP THIS IS TIMED WILL MARK BRAINIEST!!!

Anastasy [175]

Answer:

It’s A hope this is right good luck

6 0

2 years ago

Given the formula,v= 4.5 + at,find the value of v when a=1.6 and t=3.5.

jarptica [38.1K]

Answer:

v=10.1

Step-by-step explanation:

v= 4.5 + at -------> equation 1

given

a=1.6

t=3.5

so we will plug in the value in equation 1

v = 4.5 + (1.6 x 3.5)

v = 4.5 + 5.6

v = 10.1

7 0

2 years ago

Rezolvati 1) -2x+1= -7 2) 6+3x= -54+x 3) 2x+9= 7 +x 4) 3-2x= -x -6 5) -3(x+1) = -6 6) -4 (x -2)=12 7) -7(x-3)= -14

Natalija [7]

<span>Răspunsurile sunt

1. x= 4
2. x = -30
3. x = -2
4. x = 9
5. x = 1
6. x = -1
7. x = 5
</span>

3 0

3 years ago