Find the number that makes the ratio equivalent to 90:18. (blank):2

On a coordinate grid, both point (−4, −1) and (2, 6) point are reflected across the y-axis. What are the coordinates of the refl

vichka [17]

The reflected points are (4, -1) and (-2, 6).

4 0

3 years ago

Change each of the following angles in degrees to angles in radians

grin007 [14]

Answer:

d: 2.6179, e: 4.1887, f: 5.2359 all in rad

Step-by-step explanation:

(d): 150°degrees

150° × π / 180°

= 0.83333333333π rad

= 2.6179 rad

(e):240°degrees

240° × π / 180°

= 1.3333333333π rad

= 4.1887 rad

(f): 300°degrees

300° × π / 180°

= 1.6666666667π rad

= 5.2359 rad

If all my stuff is correct please give me an 5/5 and a thanks. :)

7 0

3 years ago

Can someone help me with linear functions

kow [346]

I lost my dad in the war

3 0

3 years ago

Which is the standard form of y-3=2(x-4) (A) Y=-2x-5 (B)2x-5 (C)2x+y=-5 (D)2x-y=5

pantera1 [17]

y -3 = 2(x-4)

Use distributive property on the right side:

y - 3 = 2x -8

Add 3 to both sides:

y = 2x -5

The answer is A.

4 0

3 years ago

A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of

Eva8 [605]

Answer:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=250000$ represent the sample mean

$s=37500$ represent the standard deviation for the sample

$n=5$ sample size

$\mu_o =210250$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:

Null hypothesis: $\mu \leq 210250$

Alternative hypothesis: $\mu > 210250$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=5-1=4$

Conclusion

Since is a one right tailed test the p value would be:

$p_v =P(t_{4}>2.37)=0.038$

If we compare the p value and a significance level assumed $\alpha=0.1$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.

6 0

3 years ago