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3 years ago

BRAINLIESTTTT ASAP! please answer

Mathematics

1 answer:

romanna [79]3 years ago

8 0

Simply add $b^{2}$ to both sides of the equation:

$a^{2} =c^{2} + b^{2}$

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Peter and Mary take turns throwing one dart at the dartboard. Peterhits the bulls eye with probabilitypand Mary hits the bullsey

EastWind [94]

Answer:

$P(Mary) = (1 - p) * r$

Step-by-step explanation:

Given

$p \to$ Peter hits

$r \to$ Mary hits

Required

The probability that Mary wins

From the question, we understand that Peter throws the first.

If Mary is to win, then Peter has to lose his throw

Using the complement rule, the probability that Peter misses is:

$p' = 1 - p$

So, the probability that Mary wins is:

$P(Mary) = p' * r$

This gives:

$P(Mary) = (1 - p) * r$

6 0

3 years ago

Help please soon !!!!!!

stepan [7]

75.25x + 3604 = 300.50x
300.50x - 75.25x = 3604
225.25x = 3604
x = 16
Both companies charged the same at 16 tons
first answer 16 tons
--------------------
300.50x = 300.50(16) = $4,808
both companies charged the same are $4,808
second answer is $4,808

5 0

3 years ago

"Three times a number increased by eleven is seventeen" translates to which of the following equations?

ololo11 [35]

3x + 11 = 17 i think

7 0

3 years ago

Read 2 more answers

The market and Stock J have the following probability distributions:

denis-greek [22]

Answer:

1) $E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

2) $E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

3) $E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

4) $E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

Step-by-step explanation:

For this case we have the following distributions given:

Probability M J

0.3 14% 22%

0.4 10% 4%

0.3 19% 12%

Part 1

The expected value is given by this formula:

$E(X)=\sum_{i=1}^n X_i P(X_i)$

And replacing we got:

$E(M) = 14*0.3 + 10*0.4 + 19*0.3 = 13.9 \%$

Part 2

$E(J)= 22*0.3 + 4*0.4 + 12*0.3 = 11.8 \%$

Part 3

We can calculate the second moment first with the following formula:

$E(M^2) = 14^2*0.3 + 10^2*0.4 + 19^2*0.3 = 207.1$

And the variance would be given by:

$Var (M)= E(M^2) -[E(M)]^2 = 207.1 -(13.9^2)= 13.89$

And the deviation would be:

$Sd(M) = \sqrt{13.89}= 3.73$

Part 4

We can calculate the second moment first with the following formula:

$E(J^2) = 22^2*0.3 + 4^2*0.4 + 12^2*0.3 =194.8$

And the variance would be given by:

$Var (J)= E(J^2) -[E(J)]^2 = 194.8 -(11.8^2)= 55.56$

And the deviation would be:

$Sd(M) = \sqrt{55.56}= 7.45$

8 0

3 years ago

Anybody know number 11??

PIT_PIT [208]

The property shown in #11 is the Associative Property of Addition.

In short, the associative property of addition states that no matter how an additive expression is grouped, the quantity will turn out the same. In this case, the answer is going to be equal whether you decide to add 5 and 8 first or 8 and 11 first.

5 0

3 years ago