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3 years ago

Blitzkrieg is the term used to describe Germany's?

Mathematics

2 answers:

Andreyy893 years ago

5 0

“lightning” warfare with tanks used to end a fight swiftly.

Temka [501]3 years ago

3 0

Answer:

use of fast and strong armed forces (APEX)

Step-by-step explanation:

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If two orders are selected, find the probability that they are both from Restaurant D. a. Assume that the selections are made w

Katena32 [7]

Answer:

The probability is

nothing.

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2 years ago

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

Isip at kilos loob ng<br> paaralan

Afina-wow [57]

Lol I actually laughed at this silly much ?

7 0

3 years ago

PLzzzzzzz helppppppppp!!!

fomenos

5. y = -3 x - 8

6. y = - 1/2 x + 1

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2 years ago

A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life

alisha [4.7K]

Answer:

The null hypothesis is $H_0: \mu \leq 10$

The alternative hypothesis is $H_1: \mu > 10$

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

$H_0: \mu \leq 10$

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

$H_1: \mu > 10$

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that $\mu = 10$

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that $n = 45, X = 10.5, s = 1.8$

Then

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}$

$t = 1.86$

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

5 0

2 years ago