Use the figure to find the angle measures.

Which statement is true regarding the functions on the

prisoha [69]

Answer:

f(2) = g(2)

General Formulas and Concepts:

Alg I

Reading a Cartesian Plane
Identifying Coordinates
Solutions of systems of equations

Step-by-step explanation:

We see from the graph that f(x) and g(x) intersect at x = 2. Therefore, the point at x = 2 would be equivalent in both graphs/be a solution to both equations.

Therefore, f(2) must equal g(2), as they intersect each other at that point and have the same value of 0.

3 0

3 years ago

6.04)The equation of the line of best fit of a scatter plot is y = −8x + 3. What is the the y-intercept? −8 −3 3 8

liberstina [14]

The y-intercept is 3...when plottibg this plot 3 first then use -8x (-8/1) as your slope form your y-intercept going both directions.

4 0

3 years ago

PLEASE HELP. I need DE AND EF!!!

AlekseyPX

Answer:

DE = about 41.843 (rounded to nearest thousandth)

EF= 34.276 (rounded)

Step-by-step explanation:

For DE, we know that the shorter side (the opposite side) is 24, while the angle across form it is 35°. We can use trigonometry to figure this out. SinФ equals the opposite side (in this case, 24) divided by the hypotenuse. Set sinФ equal to a ratio of the sides like this:

sin(35) = $\frac{24}{x}$

x represents the hypotenuse length, which we don't know; 35 is the angle measure. Next, isolate x so that the equation looks like this:

$\frac{24}{sin(35)}$ = x

You will need a calculator for the next part. (and make sure you're in degree mode!). evaluate sin(35) and divide 24 by that value. That is DE's length. DE = about 41.843 (rounded to nearest thousandth)

For EF, we can just use Pythagorean theorem now that we know the other sides' values.

EF^2 + 24^2 = DE^2

*a calculator might also be useful for this part.

EF= 34.276 (rounded)

8 0

2 years ago

The diameter of a circle is 12 yards. What is the circle's area? Use 3.14 for .

slamgirl [31]

Answer:

r= d/2 => 12/2 => 6

a= πR²

a= 3.14×{6}²

a= 113.04 square yard (yd²)

or

a= 94.51 m²

4 0

3 years ago

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