Find the slope of the line that passes through (-13, 64) and (69, 23).

What is the colon form? :)

Ann [662]

Answer and Step-by-step explanation:

I think this is correct? The directions aren't very clear/descriptive, so sorry if this is wrong.

A.

1. wins:games

2. wins:losses

3. losses:wins

4. games played:wins

5. losses:games played

B.

1. 8:12, or 2:3

2. 3:7

3. 5:3

4. 12:15, or 4:5

5. 20:25, or 4:5

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

4 0

3 years ago

TEST QUESTION PLS HELP!

Marat540 [252]

It should be f since you can subtract from both sides and divide by negative 1

8 0

3 years ago

Which is the function represented by the table?

zheka24 [161]

The answer is A, you would plug in 3 for fx so therefore the answer would be A. Then to make sure plug in the other functions.:)

5 0

3 years ago

Read 2 more answers

A test of H0: p = 0.5 versus Ha: p > 0.5 has the test statistic z = 1.15. Part A: What conclusion can you draw at the 5% sign

Nadya [2.5K]

Answer:

Part A: The null hypothesis failed to be rejected.

Part B: The null hypothesis failed to be rejected.

Step-by-step explanation:

We have an hypothesis test with null and alternative hypothesis H0: p = 0.5 versus Ha: p > 0.5, which has the test statistic z=1.15.

Part A: If the significance level is 0.05, the conclusion depends on the P-value.

If the P-value is below 5%, the null hypothesis is rejected.

The P-value for this right-tailed tes and z=1.15 is:

$P-value=P(z>1.15)=0.125$

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

Part B: In this case, the significance level is 0.01 and, as the alternative hypothesis is defined with an unequal sign, the test is two-tailed.

This changes the way we calculate the P-value, as we need to compute the two tails.

The P-value is:

$P-value=2P(z>1.15)=0.25$

The P-value is bigger than the significance level, so the effect is not significant and the null hypothesis is failed to be rejected.

5 0

3 years ago

(c). It is well known that the rate of flow can be found by measuring the volume of blood that flows past a point in a given tim

aleksklad [387]

(i) Given that

$V(R) = \displaystyle \int_0^R 2\pi K(R^2r-r^3) \, dr$

when R = 0.30 cm and v = (0.30 - 3.33r²) cm/s (which additionally tells us to take K = 1), then

$V(0.30) = \displaystyle \int_0^{0.30} 2\pi \left(0.30-3.33r^2\right)r \, dr \approx \boxed{0.0425}$

and this is a volume so it must be reported with units of cm³.

In Mathematica, you can first define the velocity function with

v[r_] := 0.30 - 3.33r^2

and additionally define the volume function with

V[R_] := Integrate[2 Pi v[r] r, {r, 0, R}]

Then get the desired volume by running V[0.30].

(ii) In full, the volume function is

$\displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

Compute the integral:

$V(R) = \displaystyle \int_0^R 2\pi K(R^2-r^2)r \, dr$

$V(R) = \displaystyle 2\pi K \int_0^R (R^2r-r^3) \, dr$

$V(R) = \displaystyle 2\pi K \left(\frac12 R^2r^2 - \frac14 r^4\right)\bigg_0^R$

$V(R) = \displaystyle 2\pi K \left(\frac{R^4}2- \frac{R^4}4\right)$

$V(R) = \displaystyle \boxed{\frac{\pi KR^4}2}$

In M, redefine the velocity function as

v[r_] := k*(R^2 - r^2)

(you can't use capital K because it's reserved for a built-in function)

Then run

Integrate[2 Pi v[r] r, {r, 0, R}]

This may take a little longer to compute than expected because M tries to generate a result to cover all cases (it doesn't automatically know that R is a real number, for instance). You can make it run faster by including the Assumptions option, as with

Integrate[2 Pi v[r] r, {r, 0, R}, Assumptions -> R > 0]

which ensures that R is positive, and moreover a real number.

5 0

2 years ago