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BigorU [14]
2 years ago
13

Ginger's puppy weighed pounds when she first took him home. The veterinarian said the puppy cannot eat dry dog food until he wei

ghs pounds. How many pounds must the puppy gain before he can eat dry dog food?
Mathematics
1 answer:
Tanya [424]2 years ago
5 0

Answer:

For someone to be able to answer this question you need to include the weights. Im trying to help but we cant help unless you give the amount of weight. :(

Step-by-step explanation:

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Explain what you would do first to simplify the expression below. Justify why, and then state the result of performing this step
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1 year ago
Nolan Walker decided to buy a used snowmobile since his credit union was offering such low interest rates. He borrowed $4,400 at
Serhud [2]

Answer:

He paid $253.09 in interest.

Step-by-step explanation:

To find how much did he pay in interest, we use the simple intrest formula, that is given by:

I = Prt

In which I is the value paid in interest, P is the money borrowed, r is the yearly interest rate and t is the time.

In our problem, we have that:

He borrowed $4,400, so P = 4,400

At 4.75% yearly. We measure the time in days, so we have to divide this value by 365. So r = \frac{0.0475}{365}.

From December 26, 2019 to February 21, 2021, there are 422 days, so t = 422.

I = Prt

I = 4,400\frac{0.0475}{365} 422

I = 253.09

He paid $253.09 in interest.

5 0
1 year ago
Resistors are labeled 100 Ω. In fact, the actual resistances are uniformly distributed on the interval (95, 103). Find the mean
Zinaida [17]

Answer:

E[R] = 99 Ω

\sigma_R = 2.3094 Ω

P(98<R<102) = 0.5696

Step-by-step explanation:

The mean resistance is the average of edge values of interval.

Hence,

The mean resistance, E[R] = \frac{a+b}{2}  = \frac{95+103}{2} = \frac{198}{2} = 99 Ω

To find the standard deviation of resistance, we need to find variance first.

V(R) = \frac{(b-a)^2}{12} =\frac{(103-95)^2}{12} = 5.333

Hence,

The standard deviation of resistance, \sigma_R = \sqrt{V(R)} = \sqrt5.333 = 2.3094 Ω

To calculate the probability that resistance is between 98 Ω and 102 Ω, we need to find Normal Distributions.

z_1 = \frac{102-99}{2.3094} = 1.299

z_2 = \frac{98-99}{2.3094} = -0.433

From the Z-table, P(98<R<102) = 0.9032 - 0.3336 = 0.5696

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2 years ago
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Answer:

The answer is 5.5

Step-by-step explanation:

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