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3 years ago

Some of these are wrong can someone help me please??!?!

Mathematics

2 answers:

stealth61 [152]3 years ago

6 0

Product is the result of two numbers multiplied, and some of your answers are not multiplied

yulyashka [42]3 years ago

5 0

One Is Right.
Two Is -13 for sum
Three Is 0 for sum
Four Is 14 for product and 9 for sum
Five Is 6 for sum
Six Is -3 for sum.
Seven Is 1 for sum
Eight Is -13 for sum
Nine Is -42 For Product
Ten Is 32 For Product And 12 For Sum
Eleven Is -63 for product and -2 for sum
Twelve Is right
Thirteen Is -13 for sum
Fourteen Is 0 For sum
Fiveteen Is right
Sixteen Is Wrong. -9 + -9 = -18
Seventeen Is 19 for sum and 88 for product
Eightteen Is right
Nineteen Is -5 for sum and -84 for product
Twenty Is -18 for sum
Remember, Adding Negative To Negative Makes The Negative Larger.
Remember, Adding Negative To Positive Makes Positive Smaller.
Remember, Adding Positive To Positive Makes Positive Greater. 
Remember, Negative Times Negative Is Positive.
Remember, Negative Times Positive Is Negative.
Remember, Positive Times Positive Is Positive.

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Let A and B be events with =PA0.7, =PB0.3, and =PA or B0.9. (a) Compute PA and B. (b) Are A and B mutually exclusive? Explain. (

kvasek [131]

Answer:

a) $P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

b) False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

c) False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

Step-by-step explanation:

For this case we have the following probabilities given:

$P(A) = 0.7, P(B) =0.7, P(A \cup B) =0.9$

Part a

We want to calculate the following probability: $P(A \cap B)$

And we can use the total probability rule given by:

$P(A \cup B) = P(A) +P(B) - P(A\cap B)$

And if we solve for $P(A \cap B)$ we got:

$P(A \cap B) = P(A) + P(B) -P(A\cup B)= 0.7+0.3-0.9 = 0.1$

Part b

False

The reason is because we don't satisfy the following relationship:

$P(A\cup B) = P(A) + P(B)$

We have that:

$0.9 \neq 0.3+0.7 =1$

Part c

False

In order to satisfy independence we need to have the following condition:

$P(A \cap B) = P(A) *P(B)$

And for this case we don't satisfy this relation since:

$0.1 \neq 0.7*0.3 = 0.21$

4 0

4 years ago

Evaluate please like pretty please

yaroslaw [1]

It would be zero because you can’t raise 0 to any positive power.

8 0

3 years ago

Please help really confused on all

ollegr [7]

$\left \{ {{x - y = 1.50} \atop {40x + 40y = 940}} \right. 
 \\ \\
*x - y = 1.50$

x = 1.50 + y

*40x + 40y = 940
40 (1.50 + y) + 40y = 940
60 + 40y +40y = 940
60 + 80y = 940
80y = 940 - 60
80y = 880
y = 880/80
y = 11

*x - y = 1.50
x - 11 = 1.50
x = 1.50 + 11
x = 12.5

x=DAVID
Y=PETER

5 0

3 years ago