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3 years ago

Any value that makes an equation true is a _____

Mathematics

2 answers:

djverab [1.8K]3 years ago

7 0

Answer: solution

Step-by-step explanation: it is a solution because it makes the equation true and it can be proven.

tamaranim1 [39]3 years ago

5 0

Any value that makes an equation true is a solution

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Anna [14]

$(x^3-8x^2+19x-9)\div(x-4)=x^2-4x+3\\\underline{-x^3+4x^2}\\=-4x^2+19x\\\underline{\ \ \ \ \ 4x^2-16x}\\.\ \ \ =3x-9\\\underline{\ \ \ \ \ -3x+12}\\.\ \ \ \ \ =\ 3$

$x^3-8x^2+19x-9=(x^2-4x+3)(x-4)+3$

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3 years ago

What is the runner’s average rate of change between the hours: 0.5 and 2? mph 1.5 and 2.5? mph Average Rate of Change

cestrela7 [59]

Answer:4mph

5mph

Slower

Step-by-step explanation:

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3 years ago

On Mother’s Day,Mr.Davis took Mrs.Davis and their daughter,Sue,to a restaurant for dinner.All three ordered the Mother’s Day Spe

meriva

Answer:

$52.08

Step-by-step explanation:

Mr. Davis got the Mother's Day Special regularly priced, which is $24.80. For Mrs. Davis, 40% off of $24.80 is $14.88. For Sue, half of $24.80 is $12.40. Once you add them add up, you get $52.08.

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3 years ago

A certain manufactured item is visually inspected by two different inspectors. When a defective item comes through the line, the

andreev551 [17]

The fraction of a defective item getting by both inspectors is $\frac{5}{100}$ = $\frac{1}{20}$

Step-by-step explanation:

Step 1; Assume that the probability of the first inspector missing a defective part is P(A) and the probability of the second inspector missing those that do get past the first inspector is P(B).

Step 2; It is given that P(A) = 0.1, we convert this into a fraction so that the final probability will be a fraction and not a decimal.

P(A) = 0.1 = $\frac{1}{10}$ .

It is given that the second inspector misses 5 out of 10 that get past the first inspector, so P(B) = $\frac{5}{10}$ .

Step 3; To calculate the probability of both inspectors missing a defective part, we multiply both the probabilities.

P(A and B happening) = P(A) × P(B) = $\frac{1}{10}$ × $\frac{5}{10}$ = $\frac{5}{100}$ = $\frac{1}{20}$ = 0.05%. So there is a 0.05% chance of both inspectors missing a defective part.

8 0

2 years ago

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 = the number of defective boards in a random sample of size, n = 25

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

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

= 0.093

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

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

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

= 0.966

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

= 1 - 0.966

= 0.034

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

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