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muminat
4 years ago
6

20 POINTS PLEASE HELP QUICKLY!!

Mathematics
1 answer:
SIZIF [17.4K]4 years ago
6 0

Answer:

They will each have saved $70.

Jessica is saving at a faster rate than Raphael.

The rate at which Jessica is adding to her savings is $4 per week more than the rate at which Raphael is adding to his savings.

Step-by-step explanation:

The formula for Jessica’s savings is

y = 10x + 20

She is saving at the rate of $10/wk .

From Raphael’s table, he started with $40 and is saving at the rate of $6/wk.

The formula for Raphael’s savings is

y = 6x +40

When the two savings are equal

10x + 20 = 6x + 40     Subtract 20 from each side

       10x = 6x + 20     Subtract 6x from each side

        4x = 20              Divide each side by 4

          x = 5

Their savings will be equal after five weeks.

=====

After five weeks , Jessica will have saved

y = 10×5 + 20  

y = 50 + 20

y = $70

Raphael has also saved $70.

=====

Jessica is saving at a faster rate than Raphael.

She is saving at the rate of $10/wk, while he is saving at the rate of $6/wk.

The rate at which Jessica is adding to her savings is $4/wk more than the rate at which Raphael is adding to his savings.

The red line in the graph shows that Jessica starts with $20. After five weeks, she has caught up to Raphael, who also has $70 after five weeks.  

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yanalaym [24]

Answer: 0.4512

Step-by-step explanation:

A bit string is sequence of bits (it only contains 0 and 1).

We assume that the  0 and 1 area equally likely to any place.

i.e. P(0)= P(1)= \dfrac{1}{2}

The length of bits : n = 10

Let X = Number of getting ones.

Then , X \sim Bin(n=10,\ p=\dfrac{1}{2})

Binomial distribution formula : P(X=x)=^nC_x p^x q^{n-x} , where p= probability of getting success in each event and q= probability of getting failure in each event.

Here , p=q=\dfrac{1}{2}

Then ,The probability that a bit string of length 10 contains exactly 4 or 5 ones.

P(X= 4\ or\ 5)=P(x=4)+P(x=5)\\\\=^{10}C_4(\dfrac{1}{2})^{10}+^{10}C_4(\dfrac{1}{2})^{10}

=\dfrac{10!}{4!6!}(\dfrac{1}{2})^{10}+\dfrac{10!}{5!5!}(\dfrac{1}{2})^{10}

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=(\dfrac{1}{2})^{10}(210+252)

=(0.0009765625)(462)

=0.451171875\approx0.4512

Hence, the  probability that a bit string of length 10 contains exactly 4 or 5 ones is 0.4512.

3 0
4 years ago
Sorry if its hard to see...
frozen [14]

Answer:

its 13 if i saw the numbers right

Step-by-step explanation:

evaluate then you get the below

1-(-4)x(-3)+12/(-6)

___________

-1

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-(1-(4)x(-3)+12/(6))

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-(1+4x(-3)-2)

multiply and get

-(1-12-2)

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4 years ago
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3 years ago
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scoundrel [369]

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Step-by-step explanation:

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5 0
4 years ago
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