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1 year ago

How can we use the distributive property to find an expression equivalent to

Mathematics

2 answers:

koban [17]1 year ago

7 0

Answer: multiply 10 to both terms and subrtact products

Step-by-step explanation:

10(5 - 2x)

multipy 10 to both terms

50 20x

subtract products

50 - 20x

nordsb [41]1 year ago

4 0

Answer:

1. subtract 2×from get 3.×= 682+ 4672

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(8 1/2) - (3 3/4) = 4.75

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If ItsAkelia (me) had 1M subs and M00dybear had 1M who had more?

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Demand The demand function for a product is given by

stira [4]

Answer:

$x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}$

a) $x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027$

b) $x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294$

Step-by-step explanation:

For this case we have the following function:

$P= 8000 (1- \frac{5}{5 +e^{-0.002 x}})$

We can solve for x like this. First we can reorder the expression like this:

$\frac{P}{8000} = 1- \frac{5}{5+e^{-0.002x}}$

$\frac{5}{5+e^{-0.002x}} = 1 -\frac{P}{8000} = \frac{8000-P}{8000}$

$\frac{40000}{8000-P} = 5 + e^{-0.002x}$

Now we can apply natura log on both sids and we got:

$ln[\frac{40000}{8000-P} -5] = ln e^{-0.002x}$

$ln [\frac{5P}{8000-P}] = -0.002x$

And if we solve for x we got:

$x= -\frac{ln [\frac{5P}{8000-P}]}{0.002}$

Part a

For this case we can replace P = 200 and see what we got for x like this:

$x= -\frac{ln [\frac{5*200}{8000-200}]}{0.002} =1027.062 \approx 1027$

Part b

For this case we can replace P = 800 and see what we got for x like this:

$x= -\frac{ln [\frac{5*800}{8000-800}]}{0.002} =293.893 \approx 294$

4 0

2 years ago

Carlos conducts a study comparing the performance of individuals on the first quiz in each of two classes taken in the first ter

podryga [215]

Answer:

-0.46

Step-by-step explanation:

We have small sample sizes n1 = 9 and n2 = 9. $\bar{x}_{1} = 85.11$ , $\bar{x}_{2} = 86.56$ ; $s_{1} = 7.47$ , $s_{2} = 5.75$ .

$s_{p}^{2} = \frac{(n_{1}-1)s_{1}^{2}+(n_{2}-1)s_{2}^{2}}{n_{1}+n_{2}-2} = \frac{(9-1)(7.47)^{2}+(9-1)(5.75)^{2}}{9+9-2} = 44.43$

Because Carlos conducts a study comparing the performance of individuals on the first quiz in each of two classes taken in the first term of their freshman year, the hypothesis null is $H_{0}: \mu_{1}-\mu_{2} = 0$ , then, the observed t-value is

$t=\frac{(\bar{x}_{1}-\bar{x}_{2})-0}{s{p}\sqrt{1/n1+1/n2}} = \frac{-1.45}{6.67(0.47)} = -0.46$

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3 years ago