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2 years ago

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Which statements correctly describe the cosine and sine functions? Check all that apply. The cosine function increases on (90°,

180°) and (270°, 360°). The sine function increases on (0°, 90°) and (270°, 360°). The cosine function decreases on (0°, 180°). Both the cosine and sine functions have a minimum value of 0. Both the cosine and sine functions have a maximum value of 1. Both the cosine and sine functions are periodic.

1 answer:

Jet001 [13]2 years ago

4 0

Answer:

The three correct answers are B "The sine function increases on (0°, 90°) and (270°, 360°)." , E "Both the cosine and sine functions have a maximum value of 1.", and F "Both the cosine and sine functions are periodic."

Step-by-step explanation:

Hope this helps <3

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A manufacturing company produces valves in various sizes and shapes. One particular valve plate is supposed to have a tensile st

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Answer:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

$p_v =2*P(z>1.413)=0.158$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

Step-by-step explanation:

Data given and notation

$\bar X=5.0611$ represent the sample mean

$\sigma=0.2803$ represent the population standard deviation for the sample

$n=42$ sample size

$\mu_o =5$ represent the value that we want to test

$\alpha=0.1$ represent the significance level for the hypothesis 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 true mean is different from 5, the system of hypothesis would be:

Null hypothesis: $\mu = 5$

Alternative hypothesis: $\mu \neq 5$

If we analyze the size for the sample is > 30 but and we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

z-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

We can replace in formula (1) the info given like this:

$z=\frac{5.0611-5}{\frac{0.2803}{\sqrt{42}}}=1.413$

P-value

Since is a two sided test the p value would be:

$p_v =2*P(z>1.413)=0.158$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the average tensile strength is different from 5lbs/mm at 10% of signficance

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One brand of cookies is sold in three different box sizes: small, medium, and large. The small box has half the volume of the me

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Answer:

a

Step-by-step explanation:

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n his pocket, Hamid has $2.95 in dimes and quarters. If there are 16 coins in total, which system represents the number of dimes

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X=dimes y=quarters

0.1x + 0.25y = 2.95
x+y = 16

from the second equation, x = 16 - y and thus 0.1 (16-y) +0.25y = 2.95 ==> y=9. Finally, x + 9 = 16 ==> x=7

7 dimes, 9 quarters

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4.52 Random variables X and Y follow a joint distribution f(x, y) = 2, 0 < x ≤ y < 1, 0, otherwise. Determine the correlat

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First, you'll need to find the marginal distributions of $X,Y$ . By the law of total probability,

$P(X=x)=\displaystyle\sum_yP(X=x)P(Y=y)$

which translates to

$f_X(x)=\displaystyle\int_x^1f_{X,Y}(x,y)\,\mathrm dy=\begin{cases}2(1-x)&\text{for }0$

Similarly,

$f_Y(y)=\displaystyle\int_0^yf_{X,Y}(x,y)\,\mathrm dx=\begin{cases}2y&\text{for }0$

Compute the expectations for both random variables:

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f_X(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13$

$E[Y]=\displaystyle\int_{-\infty}^\infty y\,f_Y(y)\,\mathrm dy=\int_0^12y^2\,\mathrm dy=\frac23$

Compute the variances and thus standard deviations:

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f_X(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16$

$\implies V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}\implies\sqrt{V[X]}=\dfrac1{3\sqrt2}$

$E[Y^2]=\displaystyle\int_{\infty}^\infty y^2f_Y(y)\,\mathrm dy=\int_0^12y^3\,\mathrm dy=\frac12$

$\implies V[Y]=\dfrac12-\left(\dfrac23\right)^2=\dfrac1{18}\implies\sqrt{V[Y]}=\dfrac1{3\sqrt2}$

Compute the covariance:

$\operatorname{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

We have

$E[XY]=\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty xy\,f_{X,Y}(x,y)\,\mathrm dx\,\mathrm dy=\int_0^1\int_0^y2xy\,\mathrm dx\,\mathrm dy=\frac14$

and so

$\operatorname{Cov}[X,Y]=\dfrac14-\dfrac13\dfrac23=\dfrac1{36}$

Finally, the correlation:

$\operatorname{Corr}[X,Y]=\dfrac{\operatorname{Cov}[X,Y]}{\sqrt{V[X]}\sqrt{V[Y]}}=\dfrac{\frac1{36}}{\left(\frac1{3\sqrt2}\right)^2}=\dfrac12$

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