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3 years ago

I WILL GIVE THE BRAINLIST TO WHOEVER ANSWERS CORRECTLY

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

Answer:

uwedhfuugrfhjugo

Step-by-step explanation:

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Lily deposited 1,000 in her savings account that earns her 3% compounded annually. She forgot about it until 15 years later when

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1450 is what would be in her bank account after 15 years

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3 years ago

Y varies directly with x. When y is 105, x is 7. What is k

marta [7]

Answer:

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5. Find the general solution to y'''-y''+4y'-4y = 0

CaHeK987 [17]

For any equation,

$a_ny^(n)+\dots+a_1y'+a_0y=0$

assume solution of a form, $e^{yt}$

Which leads to,

$(e^{yt})'''-(e^{yt})''+4(e^{yt})'-4e^{yt}=0$

Simplify to,

$e^{yt}(y^3-y^2+4y-4)=0$

Then find solutions,

$\underline{y_1=1}, \underline{y_2=2i}, \underline{y_3=-2i}$

For non repeated real root y, we have a form of,

$y_1=c_1e^t$

Following up,

For two non repeated complex roots $y_2\neq y_3$ where,

$y_2=a+bi$

and,

$y_3=a-bi$

the general solution has a form of,

$y=e^{at}(c_2\cos(bt)+c_3\sin(bt))$

Or in this case,

$y=e^0(c_2\cos(2t)+c_3\sin(2t))$

Now we just refine and get,

$\boxed{y=c_1e^t+c_2\cos(2t)+c_3\sin(2t)}$

Hope this helps.

r3t40

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3 years ago

What is the simplified form of expression 6(4x-7)

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3 years ago

Read 2 more answers

he one‑sample t statistic from a sample of n = 23 observations for the two‑sided test of H 0 : μ = 15 versus H α : μ > 15 has

DedPeter [7]

Answer:

$t = 2.24$

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

$df=n-1=23-1=22$

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

$p_v =P(t_{(22)}>2.24)=0.01776$

And for this case we can conclude that:

$0.01 < p_v < 0.025$

And we will reject the null hypothesis at $\alpha=0.025$ since $p_v < \alpha$

Step-by-step explanation:

Data given and notation

$\bar X$ represent the mean height for the sample

$s$ represent the sample standard deviation

$n=23$ sample size

$\mu_o =15$ represent the value that we want to 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.

We need to conduct a hypothesis in order to check if the mean is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

If we analyze the size for the sample is > 30 but we don't know the population deviation so 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".

Calculate the statistic

For this case the statistic is given:

$t = 2.24$

P-value

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

$df=n-1=23-1=22$

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

$p_v =P(t_{(22)}>2.24)=0.01776$

And for this case we can conclude that:

$0.01 < p_v < 0.025$

And we will reject the null hypothesis at $\alpha=0.025$ since $p_v < \alpha$

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3 years ago