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timama [110]
4 years ago
8

Can someone plz tell me!!?

Mathematics
1 answer:
dezoksy [38]4 years ago
6 0

Answer:

I'm going with B, but i did some research.

Step-by-step explanation:

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Lea buys 20 picture frames. Some are 4 in. by 6 in. frames which cost $5.80 each, and the rest are 5 in. by 7 in. frames that co
MAXImum [283]
To find the answer we have to slowly analyze the command.
 For surely on the one side we want to have total cash spent, so 80.8$
We know that there are 2 types of frames 5,80$ and 9.60$.
We know too, that f is the number of frames 5.80$ bought.
Simply, to find the cash spent on those frames we have to just mulitply 5.80$ times f, so we obtain f*5.8.
We can say that number of bigger frames is 20-f.
So to find cash spent on bigger frames we have to multiply 9.60$ times (20-f), so
9.60*(20-f)
Now we can just add cost of all frames together
f*5.80+9.60(20-f)
And we know that is equal to 80.8, so
f*5.80+9.60(20-f)=80.8

As we can see, the answer is b.
5 0
3 years ago
A large pool of adults earning their first driver’s license includes 50% low-risk drivers, 30% moderate-risk drivers, and 20% hi
Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

                = [0.20]⁴

                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

6 0
4 years ago
If (1/a)+(1/b)=10 what is the value of a+b?
Goshia [24]
B) 2/5 

I think, if a and b are the same (coz then they would both equal 0.2)
3 0
3 years ago
Find the missing probability. P(B)=720,P(A|B)=14 ,P(A∩B)=?
Ivan

Answer:

7/80

Step-by-step explanation:

Given that: P(B) = 7 / 20, P(A|B)= 1 / 4

Bayes theorem is used to mathematically represent the conditional probability of an event A given B. According to Bayes theorem:

P(A|B)=\frac{P(A \cap B)}{P(B)}

Where P(B) is the probability of event B occurring, P(A ∩ B) is the probability of event A and event B occurring and P(A|B) is the probability of event A occurring given event B.

P(A|B)=\frac{P(A \cap B)}{P(B)}\\\\P(A \cap B)=P(A|B)*P(B)\\\\Substituting:\\\\P(A \cap B)=1/4*7/20=7/80\\\\P(A \cap B)=7/80

6 0
3 years ago
What’s the domain and range of y=x^2+2
denpristay [2]

Answer:

domain: all real numbers

range: {y element R : y>=2}

Step-by-step explanation:

7 0
4 years ago
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