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3 years ago

12

Use the distributive property to factor the expression below. 544 y - 12 x 7

1 answer:

Zielflug [23.3K]3 years ago

6 0

The answer is 4 (136y-21)

Explanation:

Factor 4 out of 544y.
4 (136y) - 12*7

Factor 4 out of -12*7.
4 (136y) + 4 (-3*7)

Factor 4 out of 4.
(136y) + 4 (-3*7)
4 (136y- 3*7)

Multiply -3 by 7.
4 (136y - 21)

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Find sum or difference.<br><br>15/16 - 3/4<br><br>A. 1<br>B. 3/4<br>C. 3/16<br>D. 9/16

NARA [144]

1.Find a number both denominators can go into.Which is 16.
(What ever is done to the denominator must be applied to the numerator).

15*1/16*1 - 3*4/4*4
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15/16 - 12/16

2. Subtract.

15 - 12 = 3/16

Answer: 3/16

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4 years ago

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3 years ago

Observe the sequence below 37,41,45,49

lawyer [7]

Answer:

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Step-by-step explanation:

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3 years ago

Find the exact length of the curve. x = 1 + 6t2, y = 4 + 4t3, 0 ≤ t ≤ 3

vovangra [49]

$x=1+6t^2\implies\dfrac{\mathrm dx}{\mathrm dt}=12t$

$y=4+4t^3\implies\dfrac{\mathrm dy}{\mathrm dt}=12t^2$

The length of the curve $C$ is given by

$\displaystyle\int_C\mathrm ds=\int_0^3\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\dfrac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

$=\displaystyle\int_0^3\sqrt{144t^2+144t^4}\,\mathrm dt$

$=\displaystyle12\int_0^3t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3 2t\sqrt{1+t^2}\,\mathrm dt$

$=\displaystyle6\int_0^3\sqrt{1+t^2}\,\mathrm d(1+t^2)$

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3 years ago

You measure 25 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 12.8

Firlakuza [10]

Answer:

Margin of error = 4.21 ounces

Step-by-step explanation:

According to the Question,

Given That, You measure 25 turtles' weights, and find they have a mean weight of 31 ounces. Assume the population standard deviation is 12.8 ounces

Therefore, Sample mean = 31 ounces , Sample size(n) = 25 , Alpha(α) = 0.10 & Population standard deviation(σ) = 12.8 ounces

Thus, Margin of error = $Z_{critical}$ × σ / √n ( $Z_{critical}$ at α=.010 is 1.645)

Putting The Values, We get

1.645 × (12.8 / √25 ) ⇒ 4.2112 ≈ 4.21

Thus, the maximum margin of error associated with a 90% confidence interval for the true population mean turtle weight is 4.21 ounces

8 0

3 years ago