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soldi70 [24.7K]
3 years ago
14

According to recent reports, Apple Inc. holds 27.6% of the Tablet Computer market. Suppose a consumer research company believes

that 27.6% is too high and undertakes a survey to see if their suspicion is supported by the sample data. A random sample was taken of tablet computer purchases and find that of 200 purchases, 52 are Apple tablet computers. The P-value is 0.3064 and the hypothesis is : H:P = 0.276 H:P < 0.276 You would conclude at a significance level of 5%.
There is statistically significant evidence that Apple Inc. holds less than 27.6% of the tablet computer market.
there is a 30.36% chance that Apple Inc. holds at least 27.6% of the Tablet computer market.
we don't have sufficient evidence to support the claim
We have sufficient evidence to suppot the claim, that Apple holds less than 27.6%
Mathematics
1 answer:
KonstantinChe [14]3 years ago
5 0

Answer:

We don't have sufficient evidence to support the claim

Step-by-step explanation:

Null hypothesis:

H_0: p = 0.276

Alternate hypothesis:

H_1: p < 0.276

The P-value is 0.3064

0.3064 probability of finding a sample proportion lower than the one found.

The p-value is higher than the significance level of 0.05, which means that there is no sufficient evidence to support the claim.

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Vika [28.1K]

Answer:

(7 + x + a)(7 - x - a)

Here you buddy hope it helps!

6 0
2 years ago
If X and Y are independent continuous positive random
Leni [432]

a) Z=\frac XY has CDF

F_Z(z)=P(Z\le z)=P(X\le Yz)=\displaystyle\int_{\mathrm{supp}(Y)}P(X\le yz\mid Y=y)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}P(X\le yz)P(Y=y)\,\mathrm dy

where the last equality follows from independence of X,Y. In terms of the distribution and density functions of X,Y, this is

F_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy

Then the density is obtained by differentiating with respect to z,

f_Z(z)=\displaystyle\frac{\mathrm d}{\mathrm dz}\int_{\mathrm{supp}(Y)}F_X(yz)f_Y(y)\,\mathrm dy=\int_{\mathrm{supp}(Y)}yf_X(yz)f_Y(y)\,\mathrm dy

b) Z=XY can be computed in the same way; it has CDF

F_Z(z)=P\left(X\le\dfrac zY\right)=\displaystyle\int_{\mathrm{supp}(Y)}P\left(X\le\frac zy\right)P(Y=y)\,\mathrm dy

F_Z(z)\displaystyle=\int_{\mathrm{supp}(Y)}F_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Differentiating gives the associated PDF,

f_Z(z)=\displaystyle\int_{\mathrm{supp}(Y)}\frac1yf_X\left(\frac zy\right)f_Y(y)\,\mathrm dy

Assuming X\sim\mathrm{Exp}(\lambda_x) and Y\sim\mathrm{Exp}(\lambda_y), we have

f_{Z=\frac XY}(z)=\displaystyle\int_0^\infty y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=\frac XY}(z)=\begin{cases}\frac{\lambda_x\lambda_y}{(\lambda_xz+\lambda_y)^2}&\text{for }z\ge0\\0&\text{otherwise}\end{cases}

and

f_{Z=XY}(z)=\displaystyle\int_0^\infty\frac1y(\lambda_xe^{-\lambda_xyz})(\lambda_ye^{\lambda_yz})\,\mathrm dy

\implies f_{Z=XY}(z)=\lambda_x\lambda_y\displaystyle\int_0^\infty\frac{e^{-\lambda_x\frac zy-\lambda_yy}}y\,\mathrm dy

I wouldn't worry about evaluating this integral any further unless you know about the Bessel functions.

6 0
3 years ago
which ordered pair (x,y) is a solution to the given systems of linear equations? 3x+3y=-3 and -2x+y=2
Ulleksa [173]
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now eliminate x's
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2x+2y=-2
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0x+3y=0

3y=0
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x+y=-1
x+0=-1
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(x,y)
(-1,0)
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3 years ago
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C (mb) 91/323.............
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Brianna jogs 6 laps around a track each morning. The track is 1/4 mile around how many yards does Brianna jog each morning?​
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Answer:

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Step-by-step explanation:

6 times 1/4 equals 1.5. miles to yards are 1760 to 1 mile then multiple 1.5 times 1760 equaly 2640

4 0
3 years ago
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