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3 years ago

If ( 43.65 ) ( 8.79 ) / x = ( 0.4365 ) ( 87.9 ) then value of x is:

Mathematics

1 answer:

strojnjashka [21]3 years ago

5 0

Answer:

(d) 10

Step-by-step explanation:

Multiply by x and divide by its coefficient:

(43.65)(8.79) = (0.4365)(87.9)x

(43.65)(8.79)/((0.4365)(87.9)) = x

At this point, any calculator can give you the answer. It is, perhaps, more satisfying to work out the answer without a calculator.

x = (43.65)/(0.4365) × (8.79)/(87.9)

In the first quotient, the numerator is 100 times the denominator; in the second, the denominator is 10 times the numerator.

x = (100) × (1/10) = 100/10

x = 10

_____

Moving the decimal point to the right 1 place multiplies the numerical value by 10.

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The formula for the volume of a pyramid is V = Bh. Express h in terms of B and V.<br> 0<br> 0<br> =

strojnjashka [21]

Answer:

h = $\frac{V}{B}$

Step-by-step explanation:

Given

V = Bh ( isolate h by dividing both sides by B )

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3 years ago

Read 2 more answers

The sum of two numbers is -9. The difference of the two numbers is -41. What are the two numbers?

Nataly [62]

Answer:

the two numbers are 16 and -25

Step-by-step explanation:

Let the two numbers be x and y

x+y=-9

x-y=-41

By substitution

y=-9-x

x-(-9-x)=-41

x+9+x=-41

2x=-41-9

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x=-25

y=-9-(-25)

y=16

6 0

3 years ago

The prior probabilities for events A1 and A2 are P(A1) = 0.20 and P(A2) = 0.80. It is also known that P(A1 ∩ A2) = 0. Suppose P(

Umnica [9.8K]

Answer:

(a) $A_1$ and $A_2$ are indeed mutually-exclusive.

(b) $\displaystyle P(A_1\; \cap \; B) = \frac{1}{20}$ , whereas $\displaystyle P(A_2\; \cap \; B) = \frac{1}{25}$ .

(d) $\displaystyle P(A_1 \; |\; B) \approx \frac{5}{9}$ , whereas $P(A_1 \; |\; B) = \displaystyle \frac{4}{9}$

Step-by-step explanation:

$P(A_1 \; \cap \; A_2) = 0$ means that it is impossible for events $A_1$ and $A_2$ to happen at the same time. Therefore, event $A_1$ and $A_2$ are mutually-exclusive.

By the definition of conditional probability:

$\displaystyle P(B \; | \; A_1) = \frac{P(B \; \cap \; A_1)}{P(B)} = \frac{P(A_1 \; \cap \; B)}{P(B)}$ .

Rearrange to obtain:

$\displaystyle P(A_1 \; \cap \; B) = P(B \; |\; A_1) \cdot P(A_1) = 0.25 \times 0.20 = \frac{1}{20}$ .

Similarly:

$\displaystyle P(A_2 \; \cap \; B) = P(B \; |\; A_2) \cdot P(A_2) = 0.80 \times 0.05 = \frac{1}{25}$ .

Note that:

$\begin{aligned}P(A_1 \; \cup \; A_2) &= P(A_1) + P(A_2) - P(A_1 \; \cap \; A_2) = 0.20 + 0.80 = 1\end{aligned}$ .

In other words, $A_1$ and $A_2$ are collectively-exhaustive. Since $A_1$ and $A_2$ are collectively-exhaustive and mutually-exclusive at the same time:

$\displaystyle P(B) = P(B \; \cap \; A_1) + P(B \; \cap \; A_2) = \frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ .

By Bayes' Theorem:

$\begin{aligned} P(A_1 \; |\; B) &= \frac{P(B \; | \; A_1) \cdot P(A_1)}{P(B)} \\ &= \frac{0.25 \times 0.20}{9/100} = \frac{0.05 \times 100}{9} = \frac{5}{9}\end{aligned}$ .

Similarly:

$\begin{aligned} P(A_2 \; |\; B) &= \frac{P(B \; | \; A_2) \cdot P(A_2)}{P(B)} \\ &= \frac{0.05 \times 0.80}{9/100} = \frac{0.04 \times 100}{9} = \frac{4}{9}\end{aligned}$ .