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3 years ago

I need help pls pls pls

Mathematics

1 answer:

TEA [102]3 years ago

6 0

Answer:

Mantle

Step-by-step explanation:

I think its mantle

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What are the coordinates of the point on the directed line segment from (-6, -3) to

melamori03 [73]

Given:

A point divides a directed line segment from (-6, -3) to (5,8) into a ratio of 6 to 5.

To find:

The coordinates of that point.

Solution:

Section formula: If point divides a line segment in m:n, then the coordinates of that point are

$Point=\left(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n}\right)$

A point divides a directed line segment from (-6, -3) to (5,8) into a ratio of 6 to 5. Using section formula, we get

$Point=\left(\dfrac{6(5)+5(-6)}{6+5},\dfrac{6(8)+5(-3)}{6+5}\right)$

$Point=\left(\dfrac{30-30}{11},\dfrac{48-15}{11}\right)$

$Point=\left(\dfrac{0}{11},\dfrac{33}{11}\right)$

$Point=\left(0,3\right)$

Therefore, the coordinates of the required point are (0,3).

3 0

3 years ago

Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test

BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

$\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}$

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

$\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}$

- Applying the formula, we have:

$\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\ \\ t=\frac{5}{3.1752} \\ \\ t\approx1.5747 \end{gathered}$

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

$\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}$

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0

1 year ago

9.6 *10 ^-4 in standard notation

8_murik_8 [283]

9.6 x 10^-4 = 0.00096

5 0

3 years ago

Fred bought 4 liters of liquid laundry detergent 3,250 milliliters of fabric softener

Harrizon [31]

3.250 liters of softener because you have to jump three times according to the ladder method.

8 0

3 years ago

Read 2 more answers

the volume v of a right circular cylinder of radius r and heigh h is V = pi r^2 h 1. how is dV/dt related to dr/dt if h is const

laiz [17]

In general, the volume

$V=\pi r^2h$

has total derivative

$\dfrac{\mathrm dV}{\mathrm dt}=\pi\left(2rh\dfrac{\mathrm dr}{\mathrm dt}+r^2\dfrac{\mathrm dh}{\mathrm dt}\right)$

If the cylinder's height is kept constant, then $\dfrac{\mathrm dh}{\mathrm dt}=0$ and we have

$\dfrac{\mathrm dV}{\mathrm dt}=2\pi rh\dfrac{\mathrm dt}{\mathrm dt}$

which is to say, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ are directly proportional by a factor equivalent to the lateral surface area of the cylinder ( $2\pi r h$ ).

Meanwhile, if the cylinder's radius is kept fixed, then

$\dfrac{\mathrm dV}{\mathrm dt}=\pi r^2\dfrac{\mathrm dh}{\mathrm dt}$

since $\dfrac{\mathrm dr}{\mathrm dt}=0$ . In other words, $\dfrac{\mathrm dV}{\mathrm dt}$ and $\dfrac{\mathrm dh}{\mathrm dt}$ are directly proportional by a factor of the surface area of the cylinder's circular face ( $\pi r^2$ ).

Finally, the general case ( $r$ and $h$ not constant), you can see from the total derivative that $\dfrac{\mathrm dV}{\mathrm dt}$ is affected by both $\dfrac{\mathrm dh}{\mathrm dt}$ and $\dfrac{\mathrm dr}{\mathrm dt}$ in combination.

8 0

2 years ago