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3 years ago

How do you solve this with order of operation?

Mathematics

1 answer:

Wittaler [7]3 years ago

7 0

The order order operations is "PEMDAS" Which means you would solve anything with in P: Parenthesis first, anything with E: exponents second, Do M:multiplication/D:division next, and leave A:addition/S:subtraction for last.

I hope this helps, but I need to know the equation or expression to the problem to walk you through it.

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1-116 1.2.6

ankoles [38]

Answer:

A. 30 adults; B. 24 not dolls

Step-by-step explanation:

$\begin{array}{rcl}\text{No. of adults} & = & \text{No. of visitors} \times \text{Fraction who are adults}\\& = & \text{100 visitors} \times \dfrac{\text{3 adults}}{\text{10 visitors}} \\& = & \textbf{30 adults}\\\end{array}$

If five-eighths of the prizes were dolls, then three-eighths of the prizes were not dolls.

$\begin{array}{rcl}\text{No. not dolls} & = & \text{No. of prizes} \times \text{Fraction not dolls}\\& = & \text{64 prizes} \times \dfrac{\text{3 not dolls}}{\text{8 prizes}} \\& = & \textbf{24 not dolls}\\\end{array}$

3 0

3 years ago

Solve the following proportion. P/6 = 7/8 A: P=5.25 B: P=8 C: P=10.5 D: P=42

ValentinkaMS [17]

Answer:

Step-by-step explanation:

given $\frac{P}{6}$ = $\frac{7}{8}$

using the method of cross- multiplication then

8P = 42 ( divide both sides by 8 )

P = $\frac{42}{8}$ = 5.25

4 0

3 years ago

Para calcular 1/7 de 2/5

dalvyx [7]

Answer: I'm willing to help you but, I don't get what your trying to ask.

4 0

3 years ago

Suppose that bugs are present in 1% of all computer programs. A computer de-bugging program detects an actual bug with probabili

lawyer [7]

Answer:

(i) The probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii) The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

Step-by-step explanation:

Denote the events as follows:

B = bugs are present in a computer program.

D = a de-bugging program detects the bug.

The information provided is:

$P(B) =0.01\\P(D|B)=0.99\\P(D|B^{c})=0.02$

(i)

The probability that there is a bug in the program given that the de-bugging program has detected the bug is, P (B | D).

The Bayes' theorem states that the conditional probability of an event E given that another event X has already occurred is:

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

Use the Bayes' theorem to compute the value of P (B | D) as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|B^{c})P(B^{c})}=\frac{(0.99\times 0.01)}{(0.99\times 0.01)+(0.02\times (1-0.01))}=0.3333$

Thus, the probability that there is a bug in the program given that the de-bugging program has detected the bug is 0.3333.

(ii)

The probability that a bug is actually present given that the de-bugging program claims that bug is present is:

P (B|D) = 0.3333

Now it is provided that two tests are performed on the program A.

Both the test are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is:

P (Bugs are actually present | Detects on both test) = P (B|D) × P (B|D)

$=0.3333\times 0.3333\\=0.11108889\\\approx 0.1111$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on both the first and second tests is 0.1111.

(iii)

Now it is provided that three tests are performed on the program A.

All the three tests are independent of each other.

The probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is:

P (Bugs are actually present | Detects on all 3 test)

= P (B|D) × P (B|D) × P (B|D)

$=0.3333\times 0.3333\times 0.3333\\=0.037025927037\\\approx 0.037$

Thus, the probability that the bug is actually present given that the de-bugging program claims that bugs are present on all three tests is 0.037.

4 0

3 years ago

What’s the domain and range for (2,4) (5,3) (-1,-4) (0,9) (-3,1)

Citrus2011 [14]

Step-by-step explanation: In this problem, we're asked to state the domain and range for the following relation.

First of all, a relation is just a set of ordered pairs like you see in this problem. The domain is the set of all x-coordinates for those ordered pairs. So in this case the domain or D is {2, 5, -1, 0, -3}.

The range is the set of all y-coordinates for those ordered pairs. So in this case our range or R is {4, 3, -4, 9, 1}.

3 0

3 years ago