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3 years ago

13

A rectangular gift box has dimensions that can be represented as shown in the figure. The volume of the box is 56w cubic inches.

Which of the following is not a dimension of the box?

1 answer:

suter [353]3 years ago

5 0

Answer:

Step-by-step explanation:

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Please help me answer this !!!!

Studentka2010 [4]

Answer:

d = 101

Step-by-step explanation:

t = 11

11 x 11 = 121

-20 + 121 = 101

check: 101- 121 = -20 (correct)

4 0

3 years ago

What’s (-4)(0) -4? And how to do it

Ierofanga [76]

Answer:

-4

Step-by-step explanation:

Use distributive property:

(-4)(0) -4 =

0 - 4 =

-4

Hope this helps! Please mark as brainliest

7 0

4 years ago

Read 2 more answers

Your gym membership costs $30 to join and $25 per month. Which equation gives the total cost, y, for a membership that lasts x m

laiz [17]

(30+ 25)+(25 x 12) = then put your answer but this is my way of doing it.

6 0

3 years ago

Combining like terms with negative coefficients and distribution<br> -4(2x-6)-5(3x+8) ?

Ulleksa [173]

Answer:

<em>-23x-16</em>

Step-by-step explanation:

-4(2x-6)-5(3x+8)

=-8x+24-15x-40

=-23x-16

4 0

4 years ago

Suppose you flip a fair coin N times. Let random variable h be the number of heads that occur. Use the normal approximation to e

Mariana [72]

Answer:

If n = 1000000, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If n = 10400, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If N = 102, then

$P(h < 40 or h > 60) = 1-P(40$

Step-by-step explanation:

Since the coin is fair, then the probability that a filp is heads is 1/2. Given N tries, the amount of heads can be approximated with a Normal distribution with mean μ = N *1/2 = N/2 and standard deviation σ = √(N*1/2 * 1/2) = √N/ 2

The density function of that random variable is given by de following formula

$f_X(k) = \frac{1}{\sqrt{2\pi} * \sigma} e^{\frac{-(k-\mu)^2}{2\sigma^2} } = \frac{2}{\sqrt{2\pi N}} e^{\frac{-2(k-N/2)^2}{N} }$

If n = 1000000, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If n = 10400, then

$P(h E [495000, 505000]) = \int\limits^{50500}_{495000} {\frac{2}{\sqrt{2000000\pi}} e^{\frac{-(k-500000)^2}{500000} }} \, dk$

If N = 102, then

$P(h < 40 or h > 60) = 1-P(40$

4 0

3 years ago