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3 years ago

Which of the following choices correctly names angles 2 and 3 If ba bd the value of

Mathematics

1 answer:

Maslowich3 years ago

3 0

Question 1:
The answer is D, vertical angles.
Vertical angles means that 2 angles are opposite from each other for 2 lines overlapping each other. The value of those 2 angles should be same.

Question 2:
If 2 lines are Perpendicular to each other, their angle should add up to 90°(i.e. Right angle)
So, this means that angle ABC and angle CBD should add up to 90°.
Before finding the value of angle CBD, we have to find the value of x.
4x = 90° - 30°
4x = 60°
X = 15°
Therefore, just substitute x = 15° into angle CBD.
Angle CBD = 4 x 15°
=60°

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A bag of chips states that it contains 24 oz of chips. A sample of 40 bags finds a mean of 23.7 and a standard deviation of 0.2.

PIT_PIT [208]

Answer:

a) Null hypothesis: $\mu \geq 24$

b) Alternative hypothesis: $\mu < 40$

c) $t=\frac{23.7-24}{\frac{0.2}{\sqrt{40}}}=-9.487$

d) $df=n-1=40-1=39$

$p_v =P(t_{39}$

e) Since the p value is almost 0 we have enough evidence to reject the null hypothesis at the significance level of 0.05

f) Since we reject the null hypothesis we have enough evidence to conclude the true mean is significantly less than 24 ounces

Step-by-step explanation:

Information provided

$\bar X=23.7$ represent the sample mean for the weigth in ounces

$s=0.2$ represent the standard deviation

$n=40$ sample size

$\mu_o =24$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

For this case the null hypothesis is:

Null hypothesis: $\mu \geq 24$

Part b

And the alternative would be given by the claim:

Alternative hypothesis: $\mu < 40$

Part c

We don't know the population deviation so then the statistic would be given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

And replacing the data given we got:

$t=\frac{23.7-24}{\frac{0.2}{\sqrt{40}}}=-9.487$

Part d

The degrees of freedom are given by:

$df=n-1=40-1=39$

And the p value for this case would be:

$p_v =P(t_{39}$

Part e

Since the p value is almost 0 we have enough evidence to reject the null hypothesis at the significance level of 0.05

Part f

Since we reject the null hypothesis we have enough evidence to conclude the true mean is significantly less than 24 ounces

5 0

3 years ago

In a recent survey of college professors, it was found that the average amount of money spent on entertainment each week was nor

Alenkinab [10]

Answer:

0.0918

Step-by-step explanation:

We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The mean and standard deviation of average spending of sample size 25 are

μxbar=μ=95.25

σxbar=σ/√n=27.32/√25=27.32/5=5.464.

So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.

The z-score associated with average spending $102.5

Z=[Xbar-μxbar]/σxbar

Z=[102.5-95.25]/5.464

Z=7.25/5.464

Z=1.3269=1.33

We have to find P(Xbar>102.5).

P(Xbar>102.5)=P(Z>1.33)

P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)

P(Xbar>102.5)=0.5-0.4082

P(Xbar>102.5)=0.0918.

Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.

5 0

4 years ago

Write the expression as a polynomial (x+1)(x+2)(y–3)

o-na [289]

$\text{Use FOIL method:}\ (a+b)(c+d)=ac+ad+bc+bd\\\\(x+1)(x+2)(y-3)=[(x)(x)+(x)(2)+(1)(x)+(1)(2)](y-3)\\\\=(x^2+2x+x+2)(y-3)=(x^2+3x+2)(y-3)\\\\=(x^2)(y)+(x^2)(-3)+(3x)(y)+(3x)(-3)+(2)(y)+(2)(-3)\\\\=\boxed{x^2y-3x^2+3xy-9x+2y-6}$