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3 years ago

PLEASE HELP, and dont forget to show work!

Mathematics

1 answer:

weeeeeb [17]3 years ago

6 0

Answer: -rad7, +rad7

Explanation:
- Divide the equation by 6 (x^2-7=0)
- Add 7 (x^2=7)
- Square root x to find solutions (x=rad7)
- Dont forget the radical means that there are two solutions!

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Help please! I don't really understand :/

hram777 [196]

So you take 2 hours and multiply it by 5 cause you get $5 per hour. Then multiply 6.25 by 5. Add those together to get your answer!! So 5x2= 10. 6.25x5= 31.25. THE ANSWER SHOULD BE 41.25!

4 0

2 years ago

X + 25/-8 = -6 Solve for x.<br> Simplify your answer as much as possible.

balandron [24]

Step-by-step explanation:

if there is nothing missing, we have

x + 25/-8 = -6

in order to compare or add or subtract fractions, we need to bring them all to the same denominator (bottom part).

remember, integer numbers are fractions too. like here

-6 = -6/1

25/-8 = -25/8

so, how can we bring -6/1 to .../8 ?

by multiplying 1 by 8.

but we cannot multiply only the denominator by 8. otherwise we would suddenly have

-6/8

and is -6/8 = -6/1 ? no, certainly not.

to keep the original value of the fraction we have to do the same multiplication also with the numerator (top part).

so, we actually do

-6/1 × 8/8 = -48/8

with this little trick we have now .../8 to operate with, and our transformed fraction has still the same value

-6/1 = -48/8 indeed.

so, we have

x + -25/8 = -48/8

x - 25/8 = -48/8

x = -48/8 + 25/8 = -23/8

3 0

1 year 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:

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

6 0

2 years ago

The period of the function y=sin(1/4 x) is equal to

Aleonysh [2.5K]

Answer:

$8*\pi$

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2 years ago

THIS IS MY LAST ONE BROS.

Lady bird [3.3K]

Hello!
Some sample keys for the true or false three question test would be:

Keys are: False, False, False, True, True, True, False, False, True, True, True, False, False, True, False, True, False, True. False, True, True, True, True, False. Those are all the possible combinations for the answer key. (every 3 is a group). Hope this helps! Have a great day!

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2 years ago