How to make$1.05 with dimes and quarters?

Solve for the variable

liraira [26]

Answer:

y = 25

Step-by-step explanation:

$30=\frac{6}{5} y\\\\\frac{6}{5} y=30\\\\(\frac{6}{5} y)5=30*5\\\\6y=150\\\\\frac{6y=150}{6}\\\\\boxed{y=25}$

Hope this helps!

8 0

3 years ago

Which of the following expressions have a

Yuki888 [10]

A and D i think.(sry this needed to be 20 characters long)

7 0

2 years ago

PLEASE HELP ASAP 25 POINTS

Lisa [10]

We're going to go ahead and eliminate one answer out of the four. It cannot be B (the second answer) because the graph is made of dotted lines.

Dotted Lines = > or <

Solid Lines = ≤ or ≥

Next, let's focus on the straight line on the graph.

The equation for the line is

y = x - 4

Since the shaded region is below the line,

the equation will be y < x - 4

We can now eliminate answer A

Since the second equation is y < - l x - 2 l

It will be shaded below the graph since it uses the ( < ) (less than symbol)

This means the answer is C

Hope I helped, message me if you have any questions : )

8 0

3 years ago

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

3 years ago

Urgent help!! Giving out extra points for explanations

ipn [44]

Answer:

A and b r the same but b is just under x line

C would be the answer cause if it was d it be under x line

8 0

3 years ago