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3 years ago

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Joshua walked 14 feet down the hall of his house to get the door he continued in a straight line out the door and across the yar

d to the mailbox, a distance of 24 feet he came straight back across the yard 10 feet and stopped to pet his dog How far did joshua walk in total

1 answer:

AleksandrR [38]3 years ago

8 0

Answer:

Joshua walked 48 feet in total.

Step-by-step explanation:

To find the total distance we need to sum every distance that he walked. We have:

- First: Joshua walked down the hall to the door = 14 feet

- Second: He continued in a straight line out the door and across the yard to the mailbox = 24 feet

- Third: He came straight back across the yard and stopped to pet his dog = 10 feet

Hence, the total distance is:

$d_{T} = d_{1} + d_{2} + d_{3} = 14 ft + 24 ft + 10 ft = 48 ft$

Therefore, Joshua walked 48 feet in total.

I hope it helps you!

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When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%. Let

sergiy2304 [10]

Answer:

(a) P(X=3) = 0.093

(b) P(X≤3) = 0.966

(c) P(X≥4) = 0.034

(d) P(1≤X≤3) = 0.688

(e) The probability that none of the 25 boards is defective is 0.277.

(f) The expected value and standard deviation of X is 1.25 and 1.089 respectively.

Step-by-step explanation:

We are given that when circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is 5%.

Let X = <em>the number of defective boards in a random sample of size, n = 25</em>

So, X ∼ Bin(25,0.05)

The probability distribution for the binomial distribution is given by;

$P(X=r)= \binom{n}{r} \times p^{r}\times (1-p)^{n-r} ; x = 0,1,2,......$

where, n = number of trials (samples) taken = 25

r = number of success

p = probability of success which in our question is percentage

of defectivs, i.e. 5%

(a) P(X = 3) = $\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.093</u>

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

= $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}+\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $1 \times 1 \times 0.95^{25}+25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.966</u>

(c) P(X $\geq$ 4) = 1 - P(X < 4) = 1 - P(X $\leq$ 3)

= 1 - 0.966

= <u>0.034</u>

<u></u>

(d) P(1 ≤ X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)

= $\binom{25}{1} \times 0.05^{1}\times (1-0.05)^{25-1}+\binom{25}{2} \times 0.05^{2}\times (1-0.05)^{25-2}+\binom{25}{3} \times 0.05^{3}\times (1-0.05)^{25-3}$

= $25 \times 0.05^{1}\times 0.95^{24}+300 \times 0.05^{2}\times 0.95^{23}+2300 \times 0.05^{3}\times 0.95^{22}$

= <u>0.688</u>

(e) The probability that none of the 25 boards is defective is given by = P(X = 0)

P(X = 0) = $\binom{25}{0} \times 0.05^{0}\times (1-0.05)^{25-0}$

= $1 \times 1\times 0.95^{25}$

= <u>0.277</u>

(f) The expected value of X is given by;

E(X) = $n \times p$

= $25 \times 0.05$ = 1.25

The standard deviation of X is given by;

S.D.(X) = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{25 \times 0.05 \times (1-0.05)}$

= <u>1.089</u>

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