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ahrayia [7]
2 years ago
12

Jillian's puppy weighed 16 pounds at the age of 2 months. It weighed 60 pounds at the age of 8 months. What is the percent chang

e in the puppy's weight?
Mathematics
1 answer:
mel-nik [20]2 years ago
6 0

Answer:the percent change in the puppy's weight= 275%

Step-by-step explanation:

Jillian's puppy Weight  at 2 months  =16 pounds

Jillian's puppy Weight  at 8 months  =60 pounds

percent change in the puppy's weight= change in weight / Old weight  x 100

(60 pounds -16pounds ) / 16  x 100

44/16 x 100  = 275%

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Which situation represents the equation: 5x - 18 = 42
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The correct answer is C. She makes an amount of money per hour x, then she works five hours so 5x, she then spends 18 on pizza so 5x-18, and then has 42 left over. 5x-18=42
3 0
3 years ago
How to solve for 93,94, and 95
kvasek [131]

Answer:

93- not a triangle

94- right triangle

95- I dont know sorry!


3 0
3 years ago
A researcher determines that students are active about 60 + 12 (M + SD) minutes per day. Assuming these data are normally distri
rjkz [21]

Answer:

The correct option is (b).

Step-by-step explanation:

If X \sim N (µ, σ²), then Z=\frac{X-\mu}{\sigma}, is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, Z \sim N (0, 1).

The distribution of these z-variate is known as the standard normal distribution.

The mean and standard deviation of the active minutes of students is:

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<em>σ </em> = 12 minutes

Compute the <em>z</em>-score for the student being active 48 minutes as follows:

Z=\frac{X-\mu}{\sigma}=\frac{48-60}{12}=\frac{-12}{12}=-1.0

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The correct option is (b).

4 0
3 years ago
How do I solve for h?
inessss [21]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco
Elza [17]

~~~~~~ \textit{Continuously Compounding Interest Earned Amount} \\\\ A=Pe^{rt}\qquad \begin{cases} A=\textit{accumulated amount}\dotfill & \$5150\\ P=\textit{original amount deposited}\dotfill & \$4500\\ r=rate\to 7.5\%\to \frac{7.5}{100}\dotfill &0.075\\ t=years \end{cases}

5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}

4 0
1 year ago
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