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3 years ago

It takes Rebekka 1/2 hour to jog 1 1/4 miles.

Mathematics

2 answers:

MaRussiya [10]3 years ago

6 0

D because we have divide to find the unit rate.

Oduvanchick [21]3 years ago

5 0

D because it correct and I’m smart I think

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Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce

34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

$P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)$

$= 1 - P(z \leq -2)$

Calculation the value from standard normal z table, we have,

$P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%$

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

$P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 23.7) =0.159= 15.9\%$

15.9% of of the boxes weigh less than 23.7 ounces.

7 0

3 years ago

Please help! in a hurry!

Over [174]

Answer: m∠W = 56

because UVWX is a rhombus

=> UV // WX

=> 2y + 2y - 68 = 180

<=> 4y = 248

<=> y = 62

=> m∠W = 2.62 - 68 = 56

Step-by-step explanation:

7 0

3 years ago

Third degree, with zeros of -3,-2, and 1, and passes through the point (2,11)<br><br> P(x)=

Soloha48 [4]

Answer:

$P(x)=0.55(x+3)(x+2)(x-1)$

Step-by-step explanation:

If a polynomial function is third degree function and has zeros at -3,-2, and 1, then its equation is

$P(x)=a(x-(-3))(x-(-2))(x-1)\\ \\P(x)=a(x+3)(x+2)(x-1)$

It passes through the point (2,11), so the coordinates of the point satisfy the equation of the function:

$11=a(2+3)(2+2)(2-1)\\ \\11=a\cdot 5\cdot 4\cdot 1\\ \\20a=11\\ \\a=\dfrac{11}{20}=0.55\\ \\P(x)=0.55(x+3)(x+2)(x-1)$

6 0

3 years ago

Abdul has bought 30 pounds of dog food. He feeds his dog

Cerrena [4.2K]

Answer: I think its 45 days

Step-by-step explanation: hope this helps

6 0

3 years ago

The temperature, H, in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation. H = 70 + 1

Alex73 [517]

Answer:

The temperature a t = 0 is 190 °F

The temperature a t = 1 is 100 °F

The temperature a t = 2 is 77.5 °F

It takes 1.5 hours to take the coffee to cool down to 85°F

It takes 2.293 hours to take the coffee to cool down to 75°F

Step-by-step explanation:

We know that the temperature in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation:

$H(t)=70+120(\frac{1}{4})^t$

a) To find the temperature a t = 0 you need to replace the time in the equation:

$H(0)=70+120(\frac{1}{4})^0\\H(0)=70+120\cdot 1\\H(0) = 70+120\\H(0)=190 \:\°F$

b) To find the temperature after 1 hour you need to:

$H(1)=70+120(\frac{1}{4})^1\\H(1)=70+120(\frac{1}{4})\\H(1) = 70+30\\H(1)=100 \:\°F$

c) To find the temperature after 2 hours you need to:

$H(2)=70+120(\frac{1}{4})^2\\H(2)=70+120(\frac{1}{16})\\H(2) = 70+\frac{15}{2} \\H(2)=77.5 \:\°F$

d) To find the time to take the coffee to cool down $85 \:\°F$ , you need to:

$85 = 70+120(\frac{1}{4})^t\\70+120\left(\frac{1}{4}\right)^t=85\\70+120\left(\frac{1}{4}\right)^t-70=85-70\\120\left(\frac{1}{4}\right)^t=15\\\frac{120\left(\frac{1}{4}\right)^t}{120}=\frac{15}{120}\\\left(\frac{1}{4}\right)^t=\frac{1}{8}$

$\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)$

$\ln \left(\left(\frac{1}{4}\right)^t\right)=\ln \left(\frac{1}{8}\right)$

$\mathrm{Apply\:log\:rule}=\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)$

$t\ln \left(\frac{1}{4}\right)=\ln \left(\frac{1}{8}\right)$

$t=\frac{\ln \left(\frac{1}{8}\right)}{\ln \left(\frac{1}{4}\right)}\\t=\frac{3}{2} = 1.5 \:hours$

e) To find the time to take the coffee to cool down $75 \:\°F$ , you need to:

$75=70+120\left(\frac{1}{4}\right)^t\\70+120\left(\frac{1}{4}\right)^t=75\\70+120\left(\frac{1}{4}\right)^t-70=75-70\\120\left(\frac{1}{4}\right)^t=5\\\left(\frac{1}{4}\right)^t=\frac{1}{24}$

$\ln \left(\left(\frac{1}{4}\right)^t\right)=\ln \left(\frac{1}{24}\right)\\t\ln \left(\frac{1}{4}\right)=\ln \left(\frac{1}{24}\right)\\t=\frac{\ln \left(24\right)}{2\ln \left(2\right)} \approx = 2.293 \:hours$

3 0

3 years ago