<img src="https://tex.z-dn.net/?f=%5Csf%20-9%288-2h%29" id="TexFormula1" title="\sf -9(8-2h)" alt="\sf -9(8-2h)" align="absmiddl

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Veronika [31]

2 years ago

e" class="latex-formula">

Mathematics

2 answers:

grigory [225]2 years ago

4 0

Answer:

$-72+18h$

Step-by-step explanation:

$-9(8-2h)$

$\mathrm{Multiply}-9\:\mathrm{to\:the\:values\:in\:the\:parenthesis}$

$-9(8-2h)$

$-9\cdot8=-72$

$-9\cdot-2=18$

$=-72+18h$

coldgirl [10]2 years ago

3 0

Answer:

$- 9(8 - 2h) \\ by \: distributive \: property \\ = - 9 \times 8 - ( - 9) \times ( - 2h) \\ = - 72 - 18h$

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Natalija [7]

$3(4x - 2) = 12x \\ 12x - 6 = 12x \\ 12x - 12x = 6 \\ 0 = 6$
The answer is no solution

3 0

3 years ago

Dont know how to calculate it

TEA [102]

Answer:

AB = 105

Step-by-step explanation:

Since ∠ A and ∠ B are equal then Δ ABC is isosceles with 2 congruent legs.

AC = BC , substitute values

6x - 7 = 5x + 7 ( subtract 5x from both sides )

x - 7 = 7 ( add 7 to both sides )

x = 14

Then

AB = 7x + 7 = 7(14) + 7 = 98 + 7 = 105

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2 years ago

"is it appropriate to use the normal approximation for the sampling distribution of"

Katyanochek1 [597]

Answer: Normal approximation can be used for discrete sampling distributions, such as Binomial distribution and Poisson distribution if certain conditions are met.

Step-by-step explanation: We will give conditions under which the Binomial and Poisson distribitions, which are discrete, can be approximated by the Normal distribution. This procedure is called normal approximation.

1. Binomial distribution: Let the sampling distribution be the binomial distribution $B(n,p)$ , where $n$ is the number of trials and $p$ is the probability of success. It can be approximated by the Normal distribution with the mean of $np$ and the variance of $np(1-p)$ , denoted by $N(np,np(1-p))$ if the following condition is met:

$n>9\left(\frac{1-p}{p}\right)\text{ and } n>9\left(\frac{p}{1-p}\right)$

2. Poisson distribution: Let the sampling distribution be the Poisson distribution $P(\lambda)$ where $\lambda$ is its mean. It can be approximated by the Normal distribution with the mean $\lambda$ and the variance $\lambda$ , denoted by $N(\lambda,\lambda)$ when $\lambda$ is large enough, say $\lambda>1000$ (however, different sources may give different lower value for $\lambda$ but the greater it is, the better the approximation).