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2 years ago

Need help plz i show me how to do it plz :)

Mathematics

2 answers:

DaniilM [7]2 years ago

8 0

I’m pretty sure you multiply the sides your you find the formulas for the hight and width

Alex_Xolod [135]2 years ago

8 0

The formula for the surface area of a rectangular prism is SA= 2( (width•length)+(height•length)+(height•width) )

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URGENT PLEASE HELP (15 POINTS!!!!!!)

faust18 [17]

Answer:

connect C and E with a straightedge

Step-by-step explanation:

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2 years ago

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Mary has a discount coupon of 8% at a grocery store. She buys items worth $108 from the store. How much does Mary need to pay af

o-na [289]

Answer: Mary needs to pay $99.36 after using the discount coupon.

Step-by-step explanation:

Find 8% of the total items price and subtracted it from the original price to find the amount she needs to pay after the discount.

8% of 108 = 8.64

108 - 8.64 = 99.36

So $99.36 will be left to pay after using the coupon.

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2 years ago

Please show work on all problems

melamori03 [73]

Answer:

V a s a q u e r e r o s e l a e S e c h o a l p e r r o

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2 years ago

Professors at a local university earn an average salary of $80,000 with a standard deviation of $6,000. The salary distribution

Sladkaya [172]

Answer:

d. It is at most 25 percent.

Step-by-step explanation:

According to Chebyshev's inequality, for a given number of standard deviations, k, no more than 1/k² can be more than k standard deviations from the mean. In this situation the amount of standard deviations from the mean of the upper and lower bound of salaries are:

$U=\frac{\$92,000-\$80,000}{\$6,000}=2\\L = \frac{\$80,000-\$68,000}{\$6,000}= 2$

For k = 2, applying Chebyshev's inequality:

$P( \$68,000 \leq X \leq \$92,000)= \frac{1}{2^2} = 0.25$

Therefore, at most 25% of the salaries are less than $68,000 or more than $92,000.

7 0

2 years ago

6n + n^7 is divisible by 7 and prove it in mathematical induction<br>

kompoz [17]

Answer:

Apply induction on $n$ (for integers $n \ge 1$ ) after showing that $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, \dots,\, 6 \rbrace$ .

Step-by-step explanation:

Lemma: $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7$ for $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ .

Proof: assume that for some $j \in \lbrace 1,\, 2,\, \dots,\, 6\rbrace$ , $\genfrac{(}{)}{0}{}{7}{j}$ is not divisible by $7$ .

The combination $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is known to be an integer. Rewrite the factorial $7!$ to obtain:

$\displaystyle \begin{pmatrix}7 \\ j\end{pmatrix} = \frac{7!}{j! \, (7 - j)!} = \frac{7 \times 6!}{j!\, (7 - j)!}$ .

Note that $7$ (a prime number) is in the numerator of this expression for $\genfrac{(}{)}{0}{}{7}{j}\!$ . Since all terms in this fraction are integers, the only way for $\genfrac{(}{)}{0}{}{7}{j}$ to be non-divisible by $7\!$ is for the denominator $j! \, (7 - j)!$ of this expression to be an integer multiple of $7\!\!$ .

However, since $1 \le j \le 6$ , the prime number $\!7$ would not a factor of $j!$ . Similarly, since $1 \le 7 - j \le 6$ , the prime number $7\!$ would not be a factor of $(7 - j)!$ , either. Thus, $j! \, (7 - j)!$ would not be an integer multiple of the prime number $7$ . Contradiction.

Proof of the original statement:

Base case: $n = 1$ . Indeed $6 \times 1 + 1^{7} = 7$ is divisible by $7$ .

Induction step: assume that for some integer $n \ge 1$ , $(6\, n + n^{7})$ is divisible by $7$ . Need to show that $(6\, (n + 1) + (n + 1)^{7})$ is also divisible by $7\!$ .

Fact (derived from the binomial theorem $(\ast)$ ):

$\begin{aligned} & (n + 1)^{7} \\ &= \sum\limits_{j = 0}^{7} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] && (\ast)\\ &= \genfrac{(}{)}{0}{}{7}{0} \, n^{0} + \genfrac{(}{)}{0}{}{7}{7} \, n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right] \\ &= 1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\end{aligned}$ .

Rewrite $(6\, (n + 1) + (n + 1)^{7})$ using this fact:

$\begin{aligned} & 6\, (n + 1) + (n + 1)^{7} \\ =\; & 6\, (n + 1) + \left(1 + n^{7} + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \\ =\; & 6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right) \end{aligned}$ .

For this particular $n$ , $(6\, n + n^{7})$ is divisible by $7$ by the induction hypothesis.

$\sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]$ is also divisible by $7$ since $n$ is an integer and (by lemma) each of the coefficients $\genfrac{(}{)}{0}{}{7}{j} = (7!) / (j! \, (7 - j)!)$ is divisible by $7\!$ .

Therefore, $6\, (n + 1) + (n + 1)^{7}$ , which is equal to $6\, n + n^{7} + 7 + \sum\limits_{j = 1}^{6} \left[\genfrac{(}{)}{0}{}{7}{j}\, n^{j}\right]\right)$ , is divisible by $7$ .

In other words, for any integer $n \ge 1$ , if $(6\, n + n^{7})$ is divisible by $7$ , then $6\, (n + 1) + (n + 1)^{7}$ would also be divisible by $7\!$ .

Therefore, $(6\, n + n^{7})$ is divisible by $7$ for all integers $n \ge 1$ .

6 0

1 year ago