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Free_Kalibri [48]
3 years ago
5

Pears cost $0.92 per pound and apples cost $1.10 per pound. mr.Bonilla bought 3.75 pounds of pairs and 2.1 pounds of apples how

much did they pay for the pears and apples
Mathematics
1 answer:
eimsori [14]3 years ago
4 0

Answer:

$5.76

Step-by-step explanation:

Total price for 3.75 pounds of pears:

0.92(3.75) = $3.45

Total price for 2.10 pounds of pears:

1.10(2.10) = $2.31

Total Price:

$2.31 + $3.45 = $5.76

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SVETLANKA909090 [29]
5*3=15
15-4= 11
hope I helped:)
7 0
3 years ago
If S_1=1,S_2=8 and S_n=S_n-1+2S_n-2 whenever n≥2. Show that S_n=3⋅2n−1+2(−1)n for all n≥1.
Snezhnost [94]

You can try to show this by induction:

• According to the given closed form, we have S_1=3\times2^{1-1}+2(-1)^1=3-2=1, which agrees with the initial value <em>S</em>₁ = 1.

• Assume the closed form is correct for all <em>n</em> up to <em>n</em> = <em>k</em>. In particular, we assume

S_{k-1}=3\times2^{(k-1)-1}+2(-1)^{k-1}=3\times2^{k-2}+2(-1)^{k-1}

and

S_k=3\times2^{k-1}+2(-1)^k

We want to then use this assumption to show the closed form is correct for <em>n</em> = <em>k</em> + 1, or

S_{k+1}=3\times2^{(k+1)-1}+2(-1)^{k+1}=3\times2^k+2(-1)^{k+1}

From the given recurrence, we know

S_{k+1}=S_k+2S_{k-1}

so that

S_{k+1}=3\times2^{k-1}+2(-1)^k + 2\left(3\times2^{k-2}+2(-1)^{k-1}\right)

S_{k+1}=3\times2^{k-1}+2(-1)^k + 3\times2^{k-1}+4(-1)^{k-1}

S_{k+1}=2\times3\times2^{k-1}+(-1)^k\left(2+4(-1)^{-1}\right)

S_{k+1}=3\times2^k-2(-1)^k

S_{k+1}=3\times2^k+2(-1)(-1)^k

\boxed{S_{k+1}=3\times2^k+2(-1)^{k+1}}

which is what we needed. QED

6 0
2 years ago
The parent function of a graph is f(x) = x2. The graph shifts a units to the left and down b units. Which function models the tr
Gnesinka [82]
<span>C) y = (x + a)2 - b </span>is the answer
8 0
3 years ago
Read 2 more answers
4 people went to the movies. Each person spent $9.50 on a ticket and had the same amount of money leftover. They originally had
solong [7]

$2.50 was left for each person.

4(9.50) + x = 48

38 + x = 48

subtract 38 on both sides

x = 10

10$ was left in total by all 4 people

10/4 = 2.50

$2.50 was left over for each person

7 0
3 years ago
Let μ denote the true average radioactivity level(picocuries per liter). The value 5 pCi/L is considered thedividing line betwee
VMariaS [17]

Answer:

H0: μ = 5 versus Ha: μ < 5.

Step-by-step explanation:

Given:

μ = true average radioactivity level(picocuries per liter)

5 pCi/L = dividing line between safe and unsafe water

The recommended test here is to test the null hypothesis, H0: μ = 5 against the alternative hypothesis Ha: μ < 5.

A type I error, is an error where the null hypothesis, H0 is rejected when it is true.

We know type I error can be controlled, so safer option which is to test H0: μ = 5 vs Ha: μ < 5 is recommended.

Here, a type I error involves declaring the water is safe when it is not safe. A test which ensures that this error is highly unlikely is desirable because this is a very serious error. We prefer that the most serious error be a type I error because it can be explicitly controlled.

7 0
2 years ago
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