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3 years ago

Xy-11=5 is this a direct variation

2 answers:

6 0

A direct variation equation is in the form y = kx. So if you solve this equation will you get it in that form? No you will get y = 16/x. This is an inverse variation.

butalik [34]3 years ago

6 0

The answer will be 16/x.

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On a number line, suppose point E has a coordinate of -2, and EG = 13. What are the possible coordinates of point G

mihalych1998 [28]

Some possible coordinates for G are 11 or -15

6 0

3 years ago

What is the solution to the system of linear equations

Wittaler [7]

The answer is c. (0,2)

3 0

3 years ago

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HELP!!! An electronics company designed a cardboard box for its new line of air purifiers. The figure shows the dimensions of th

Anarel [89]

Answer:

130 in^2 net area

136 in^2 gross area (counting the 6 in^2 scrap loss for the side cutouts)

Step-by-step explanation:

top=1.5x4=6

middle=1.5x4=6

bottom=3x4=12

sides=(2x3x8)-(2x1.5x2)=42

front & back=2x8x4=64

6+6+12+42+64=130

6 0

3 years ago

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A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

SO, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is equal to 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different than 3%}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is different from 3%.

3 0

3 years ago

What is the true solution to the equation below? 2 In e^In2x -In e^In10x = In 30

stealth61 [152]

$The\ domain.\\D:x\in\mathbb{R^+}\\\\2\ln e^{\ln2x}-\ln e^{\ln10x}=\ln30\ \ \ |use\ \log_ab^n=n\log_ab\\\\2\ln2x\ln e-\ln10x\ln e=\ln30\ \ \ \ |use\ \log_aa=1\\\\2\ln2x-\ln10x=\ln30\ \ \ |use\ n\log_ab^n=\log_ab^n\\\\\ln(2x)^2-\ln10x=\ln30\\\\\ln4x^2-\ln10x=\ln30\ \ \ |use\ \log_ab-\log_ac=\log_a\dfrac{b}{c}\\\\\ln\dfrac{4x^2}{10x}=\ln30\\\\\ln\dfrac{2x}{5}=\ln30\iff\dfrac{2x}{5}=30\ \ \ \ |\cdot5\\\\2x=150\ \ \ \ |:2\\\\x=75\in D$

Answer: x = 75

8 0

3 years ago

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