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bekas [8.4K]
3 years ago
7

5x + 19.95 = 35 Solve x

Mathematics
2 answers:
solniwko [45]3 years ago
7 0

Answer:

Answer is x=3+0.05/5

Step-by-step explanation:

5x+19.95=35

We move all terms to the left:

5x+19.95-(35)=0

We add all the numbers together, and all the variables

5x-15.05=0

We move all terms containing x to the left, all other terms to the right

5x=15.05

x=15.05/5

x=3+0.05/5

I hope it's helpful!!

lora16 [44]3 years ago
3 0

Answer:

X = 3.01

Step-by-step explanation:

To solve for x, you first need to get x by itself. So you would minus 19.95 from both sides. Which would then be

5x = 15.05

Again x is not by itself so you would divide both sides by 5, the eqation would then be

x = 3.01

Which is the answer. If you want you can double check your work by putting the answer back in the eqaution.

5(3.01) + 19.95 = 35

If the statement is true that means you have the right answer

15.05 + 19.95 = 35

35 = 35

We are correct! Hope this helps with your question and how to do others like it.

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A common blood test performed on pregnant women to screen for chromosome abnormalities in the fetus measures the human chorionic
goldfiish [28.3K]

Answer:

(a) The proportion of women who are tested, get a negative test result is 0.82.

(b) The proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

Step-by-step explanation:

The Bayes' theorem states that the conditional probability of an event <em>E</em>_{i}, of the sample space <em>S,</em> given that another event <em>A</em> has already occurred is:

P(E_{i}|A)=\frac{P(A|E_{i})P(E_{i})}{\sum\liits^{n}_{i=1}{P(A|E_{i})P(E_{i})}}

The law of total probability states that, if events <em>E</em>₁, <em>E</em>₂, <em>E</em>₃... are parts of a sample space then for any event <em>A</em>,

P(A)=\sum\limits^{n}_{i=1}{P(A|B_{i})P(B_{i})}

Denote the events as follows:

<em>X</em> = fetus have a chromosome abnormality.

<em>Y</em> = the test is positive

The information provided is:

P(X)=0.04\\P(Y|X)=0.90\\P(Y^{c}|X^{c})=0.85

Using the above the probabilities compute the remaining values as follows:

P(X^{c})=1-P(X)=1-0.04=0.96

P(Y^{c}|X)=1-P(Y|X)=1-0.90=0.10

P(Y|X^{c})=1-P(Y^{c}|X^{c})=1-0.85=0.15

(a)

Compute the probability of women who are tested negative as follows:

Use the law of total probability:

P(Y^{c})=P(Y^{c}|X)P(X)+P(Y^{c}|X^{c})P(X^{c})

          =(0.10\times 0.04)+(0.85\times 0.96)\\=0.004+0.816\\=0.82

Thus, the proportion of women who are tested, get a negative test result is 0.82.

(b)

Compute the value of P (X|Y) as follows:

Use the Bayes' theorem:

P(X|Y)=\frac{P(Y|X)P(X)}{P(Y|X)P(X)+P(Y|X^{c})P(X^{c})}

             =\frac{(0.90\times 0.04)}{(0.90\times 0.04)+(0.15\times 0.96)}

             =0.20

Thus, the proportion of women who get a positive test result are actually carrying a fetus with a chromosome abnormality is 0.20.

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