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3 years ago

Need to graph y is greater than or equal to- 3x

Mathematics

1 answer:

andrew-mc [135]3 years ago

6 0

= to it I want to say it's y is 1 correct me if I am wrong

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What is the value of x that makes the given equation true? X-3x=2(4+x)

Ratling [72]

x - 3x = 2(4 + x)

-2x = 8 + 2x

-4x = 8

x = -2

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3 years ago

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Please Help!!!!!!! Meryl spends a total of $68.82 for 2 pairs of sneakers

timama [110]

2x+3-5.32=68.82

2x= 71.14

x=35.57

each sneaker costs $35.57

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3 years ago

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Find the solution set (x+1)(x+1)=0

lara31 [8.8K]

X= -1 beacuse negative one plus one is 0 and that will equal zero for both sides

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3 years ago

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The Wall Street Journal reported that automobile crashes cost the United States $162 billion annually (2008 data). The average c

ivanzaharov [21]

Answer:

$n=(\frac{2.326(500)}{120})^2 =93.928 \approx 94$

So the answer for this case would be n=94 rounded up to the nearest integer

Step-by-step explanation:

Information given

$\bar X$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma= 500$ represent the population standard deviation

n represent the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =120 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The confidence2 level is 98% or 0.98 then the significance level would be $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , the critical value for this case would be $z_{\alpha/2}=2.326$ , replacing into formula (b) we got:

$n=(\frac{2.326(500)}{120})^2 =93.928 \approx 94$

So the answer for this case would be n=94 rounded up to the nearest integer

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3 years ago

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil

Fiesta28 [93]

Answer:

The sample has not met the required specification.

Step-by-step explanation:

As the average of the sample suggests that the true average penetration of the sample could be greater than the 50 mils established, we formulate our hypothesis as follow

$H_0$ : The true average penetration is 50 mils

$H_a$ : The true average penetration is > 50 mils

Since we are trying to see if the true average is greater than 50, this is a right-tailed test.

If the level of confidence is α = 0.05 then the $z_\alpha$ score against we are comparing with, is 1.64 (this is because the area under the normal curve N(0;1) to the right of 1.64 is 0.05)