Four students, Jackson, Chloe, Lydia, and Noah, line up one behind the other. How

Order the set (-5,3,2,-7) from greatest to least

STatiana [176]

3,2,-5,-7 that is the order it should be in

8 0

1 year ago

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20 pts please help!

sweet-ann [11.9K]

Did you get an answer? I thought it was lump sum, but i don't know which option

6 0

2 years ago

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Linda Gramham's Cinnamon Sugar Graham Cupcake recipe for 36 cupcakes requires 1 1/4 cups of sugar. How much sugar is required fo

Anon25 [30]

Answer:

31.25 or 31 1/4 cups

Step-by-step explanation:

1 1/4 = 36

x = 900

=> 5/4 = 36

x = 900

Cross multiply

=> 5/4 * 900 = 36 * x

=> 4500/4 = 36x

=> 1125 = 36x

=> 1125 / 36 = 36x / 36

=> 31.25 = x

31.25 = 31 1/4

So, 31.25 or 31 1/4 cups are required for 900 cupcakes.

5 0

1 year ago

The probability that the first stage of a numerically controlled machining operation for high-rpm pistons meets specifications i

lora16 [44]

Answer:

There is an 85.5% probability that both stages meet specifications.

Step-by-step explanation:

We have these following probabilities:

A 90% probability that the first stage meets specifications.

If the first stage meets specifications, a 95% probability that the second stage also meets specifications.

What is the probability that both stages meet specifications?

This is the multiplication of these probabilities. So:

$P = 0.9*0.95 = 0.855$

There is an 85.5% probability that both stages meet specifications.

4 0

1 year ago

Which of the following is the expansion of (3c + d2)6?

vovangra [49]

Answer: Option A) $729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$ is the correct expansion.

Explanation:

on applying binomial theorem, $(a+b)^n=\sum_{r=0}^{n} \frac{n!}{r!(n-r)!} a^{n-r} b^r$

Here a=3c, $b=d^2$ and n=6,

Thus, $(3c+d^2)^6=\sum_{r=0}^{6} \frac{6!}{r!(6-r)!} (3c)^{n-r} (d^2)^r$

⇒ $(3c+d^2)^6= \frac{6!}{(6-0)!0!} (3c)^{6-0}.(d^2)^0+\frac{6!}{(6-1)!1!} (3c)^{6-1}.(d^2)^1+\frac{6!}{(6-2)!2!} (3c)^{6-2}.(d^2)^2+\frac{6!}{(6-3)!3!} (3c)^{6-3}.(d^2)^3+\frac{6!}{(6-4)!4!} (3c)^{6-4}.(d^2)^4+\frac{6!}{(6-5)!5!} (3c)^{6-5}.(d^2)^5+\frac{6!}{(6-6)!6!} (3c)^{6-6}.(d^2)^6$

⇒ $(3c+d^2)^6= \frac{6!}{(6-)!0!} (3c)^6.d^0+\frac{6!}{(5)!1!} (3c)^5.d^2+\frac{6!}{(4)!2!} (3c)^4.d^4+\frac{6!}{(6-3)!3!} (3c)^3.d^6+\frac{6!}{(2)!4!} (3c)^2.d^8+\frac{6!}{(1)!5!} (3c).d^{10}+\frac{6!}{(0)!6!} (3c)^0.d^{12}$

⇒ $(3c+d^2)^6=(3c)^6.d^0+\frac{720}{120} (3c)^5.d^2+\frac{720}{48} (3c)^4.d^4+\frac{720}{36} (3c)^3.d^6+\frac{720}{48} (3c)^2.d^8+\frac{720}{120} (3c).d^{10}+.d^{12}$

⇒ $(3c+d^2)^6=729c^6 + 1,458c^5d^2 + 1,215c^4d^4 + 540c^3d^6 + 135c^2d^8 + 18cd^{10} + d^{12}$

3 0

1 year ago

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