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romanna [79]
3 years ago
5

Help with these two problems please

Mathematics
1 answer:
Troyanec [42]3 years ago
3 0

Answer:

Can you please give me the question to see if I can answer it

Step-by-step explanation:

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What is the value of p?
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61

Step-by-step explanation:

comment if you want a step-by-step

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Veronica deposited $21 in a savings account that earns 2.4% simple interest. Which graph represents this scenario?
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Where’s the graph at?
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5 Roy wants to buy a new television for $300. Two stores
nasty-shy [4]

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Initial value for store A = 300$

Initial value for store B = 200$

Explanation: The initial value is the value of y, when x is 0. Here the x axis will represent the month, while y would represent the amount the buyer owes on that month. For store A, at month 0, the buyer owes 300$. For store B, there is an instant payment of 100$, leaving the buyer to owe 200$ from the start of the deal, hence the reason for my initial values above.

Step-by-step explanation:

Rate of Change for store A = 300 - 250 = 50

Rate of change for Store B = 50

The rate of change in this case is the same, as the rate of change is defined as how fast the output change in relation to the input. In both cases, they are required to pay 50$ each month.

6 0
3 years ago
A right triangle is drawn on a coordinate plane. Two vertices of the triangle are points R(−9,6) and T(10,−7) . The third vertex
SSSSS [86.1K]
The coordinates of point S would be (10,6), so the distance from point T to point S would be 13. :) 
4 0
3 years ago
From experience an airline knows that only 85% passengers
V125BC [204]

Answer:

P(x

Step-by-step explanation:

Let's call p the probability that a passenger shows up.

Then we know that:

p = 0.85

Then we took a sample of n = 6 passengers.

We can calculate the probability that less than 4 are presented using the binomial formula:

P(x) = \frac{n!}{x!(n-x)!}*p^x*(1-p)^{n-x}

Where x is the number of passengers that show up, n is the number of selected passengers, p is the probability that a passenger shows up.

Then we look for:

P(x

P(6) = \frac{6!}{6!(6-6)!}*0.85^6*(1-0.85)^{6-6}=0.37715

P(5) = \frac{6!}{5!(6-5)!}*0.85^5*(1-0.85)^{6-5}=0.39933

P(4) = \frac{6!}{4!(6-4)!}*0.85^4*(1-0.85)^{6-4}=0.17618

P(x

P(x

6 0
3 years ago
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