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3 years ago

14

Rene has a coupon for $3.25 off a package of name brand cookies that normally costs $7.89. The store brand cookies coats $5.58.

How much will Rene save if she uses her coupon and buys the name brand cookies instead of the store brand cookies? HELP

1 answer:

allochka39001 [22]3 years ago

7 0

She would save 94 cents buying the name brand cookies

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Help on horizontal asymptote!

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Answer:

A. None

Step-by-step explanation:

They are no horizontal asymptotes. I get helped from another website when comes to graphs and looking for asymptotes. Good luck on your exam!

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3 years ago

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Candy and Tim share a paper route. It takes Candy 85 min to deliver all the papers, and it takes Tim 65 min. How long does it ta

RSB [31]

Answer:

The time required to finish the work by both candy and Tim working together = 36.83 minute

Step-by-step explanation:

Time taken by the candy to finish the work $T_{1}$ = 85 min

Time taken by the Tim to finish the work $T_{2}$ = 65 min

Thus the time taken by both to finish the work = $\frac{T_1 \times\ T_2}{T_1 + T_2}$

Put the values of $T_{1}$ and $T_{2}$ we get ,

$T_{}$ = $\frac{85 \times\ 65}{85 + 65}$

$T_{}$ = $\frac{221}{6}$

$T_{}$ = 36.83 minute

Thus the time required to finish the work by both candy and Tim working together = 36.83 minute

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3 years ago

Do these ratios form a proportion?

ad-work [718]

Answer:

no

Step-by-step explanation:

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3 years ago

The number of people arriving at a ballpark is random, with a Poisson distributed arrival. If the mean number of arrivals is 10,

Stella [2.4K]

Answer:

a) 3.47% probability that there will be exactly 15 arrivals.

b) 58.31% probability that there are no more than 10 arrivals.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given time interval.

If the mean number of arrivals is 10

This means that $\mu = 10$

(a) that there will be exactly 15 arrivals?

This is P(X = 15). So

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 15) = \frac{e^{-10}*(10)^{15}}{(15)!} = 0.0347$

3.47% probability that there will be exactly 15 arrivals.

(b) no more than 10 arrivals?

This is $P(X \leq 10)$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-10}*(10)^{0}}{(0)!} = 0.000045$

$P(X = 1) = \frac{e^{-10}*(10)^{1}}{(1)!} = 0.00045$

$P(X = 2) = \frac{e^{-10}*(10)^{2}}{(2)!} = 0.0023$

$P(X = 3) = \frac{e^{-10}*(10)^{3}}{(3)!} = 0.0076$

$P(X = 4) = \frac{e^{-10}*(10)^{4}}{(4)!} = 0.0189$

$P(X = 5) = \frac{e^{-10}*(10)^{5}}{(5)!} = 0.0378$

$P(X = 6) = \frac{e^{-10}*(10)^{6}}{(6)!} = 0.0631$

$P(X = 7) = \frac{e^{-10}*(10)^{7}}{(7)!} = 0.0901$

$P(X = 8) = \frac{e^{-10}*(10)^{8}}{(8)!} = 0.1126$

$P(X = 9) = \frac{e^{-10}*(10)^{9}}{(9)!} = 0.1251$

$P(X = 10) = \frac{e^{-10}*(10)^{10}}{(10)!} = 0.1251$

$P(X \leq 10) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) = 0.000045 + 0.00045 + 0.0023 + 0.0076 + 0.0189 + 0.0378 + 0.0631 + 0.0901 + 0.1126 + 0.1251 + 0.1251 = 0.5831$

58.31% probability that there are no more than 10 arrivals.

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3 years ago

What is the rate of change.

Finger [1]

Answer:

-1

Step-by-step explanation:

Over 12 on the x axis it goes down 12 on the y axis so the rate of change is -1

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3 years ago