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3 years ago

Anybody know the right answer?

Mathematics

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

If we just have

$y = \cos x$

that's an amplitude of 1 and a period of $2\pi$

Let's modify this step by step.

Amplitude of 4. Four times the amplitude means a factor of 4 on the outside.

$y = 4 \cos x$

Period of π. Double the period means a factor of 2 on the inside.

$y = 4 \cos(2x)$

Horizontal shift of π/2 to the left. To the left means adding a positive number to x. The question appear ambiguous. It's not clear to me if we add to x or 2x; let's make it

$y = 4 \cos(2(x + \frac \pi 2))$

Vertical shift of 3:

$y = 3 + 4 \cos(2(x + \frac \pi 2))$

$y = 3 + 4 \cos(2x + \pi)$

Depending on the interpretation of the phase shift, that pi may be a pi/2.

Answer: $3 + 4 \cos(2x + \pi)$

You might be interested in

A group of 15 people are going to run a race. The top 8 will advance to the finals. Determine if the situation is a permutation

LUCKY_DIMON [66]

For the given situation we have a total of 259,459,200 permutations.

<h3>How many permutations are?</h3>

First, how we know that it is a permutation?

Because the order matters, we aren't only selecting 8 out of the 15 people, but these 8 selected also have an order (is not the same thing to finish the race first than fourth, for example).

Then we need to find the number of permutations, to do it, we need to find the numbers of options for each of the 8 positions.

For the first position there are 15 options.
For the second position ther are 14 options (one runner already finished).
For the third position there are 13 options.
And so on.

Then the total number of permutations (product between the numbers of options) is:

P = 15*14*13*12*11*10*9*8 = 259,459,200

If you want to learn more about permutations:

brainly.com/question/11732255

#SPJ1

8 0

2 years ago

A study of the number of business lunches that executives in the insurance and banking industries claim as deductible expenses p

gizmo_the_mogwai [7]

Answer:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

$p_v =2*P(t_{(88)}>2.604)=0.0108$

So the p value is a very low value and using any significance level for given $\alpha=0.05$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and there is enough evidence to conclude that the two means are significantly different at 5%

Step-by-step explanation:

Data given and notation

$\bar X_{1}=9.1$ represent the mean for the sample 1

$\bar X_{2}=8$ represent the mean for the sample 2

$s_{1}=1.9$ represent the sample standard deviation for the sample 1

$s_{2}=2.1$ represent the sample standard deviation for the sample 2

$n_{1}=40$ sample size for the group 1

$n_{2}=50$ sample size for the group 2

t would represent the statistic (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the mean are different , the system of hypothesis would be:

Null hypothesis: $\mu_{1} = \mu_{2}$

Alternative hypothesis: $\mu_{1} \neq \mu_{2}$

If we analyze the size for the samples both are higher than 30 and the population deviations are not given, so for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

We can replace in formula (1) the results obtained like this:

$t=\frac{9.1-8}{\sqrt{\frac{(1.9)^2}{40}+\frac{(2.1)^2}{50}}}}=2.604$

Statistical decision

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=40+50-2=88$