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Svetradugi [14.3K]
2 years ago
10

4. G.CO.10 In the figure below, p ll q. What is the value of x? *

Mathematics
1 answer:
Natali5045456 [20]2 years ago
7 0
Correct answer is C 59
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You have 15 pennies in your pocket. Of those pennies, 3 are Canadian. Suppose you pick a penny out of your pocket at random. Fin
choli [55]
12 out of 15. 12/15 = 4/5 answer is b
4 0
3 years ago
Read 2 more answers
A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig
jenyasd209 [6]

Answer:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

n=25 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

Alternative hypothesis:\mu < 20  

The statistic is given by:

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

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

7 0
3 years ago
The drive from savannah to Milwaukee takes approximately 15.5 hours how long is the drive in minutes
Anna71 [15]

Answer: the result is 930 minutes.

Step-by-step explanation:

15.5 hours to minutes has been calculated by multiplying 15.5 hours by 60

4 0
3 years ago
Write in radical form
Studentka2010 [4]

Step-by-step explanation:

I'll do one for you.

Using the formula for turning exponents into radicals

(b) {}^{ \frac{x}{y} }  =    \sqrt[y]{b} {}^{x}

where b is the base

This means that the numerator in the exponet form becomes the power under the radical in radical form and

the denominator in exponet form becomes the nth root in radical form.

For example 5,

(5) {}^{ \frac{ - 3}{5} }

That becomes in radical form

\sqrt[5]{5 {}^{ - 3} }

or if you want to write it using positive exponents

\frac{1}{ \sqrt[5]{5 {}^{3} } }

I'll do one more for you

For example 6,

11 {}^{ \frac{4}{3} }

That becomes in radical form

\sqrt[3]{11 {}^{4} }

4 0
3 years ago
in exercises 15-20 find the vector component of u along a and the vecomponent of u orthogonal to a u=(2,1,1,2) a=(4,-4,2,-2)
Nimfa-mama [501]

Answer:

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

Step-by-step explanation:

Let \vec u and \vec a, from Linear Algebra we get that component of \vec u parallel to \vec a by using this formula:

\vec u_{\parallel \,\vec a} = \frac{\vec u \bullet\vec a}{\|\vec a\|^{2}} \cdot \vec a (Eq. 1)

Where \|\vec a\| is the norm of \vec a, which is equal to \|\vec a\| = \sqrt{\vec a\bullet \vec a}. (Eq. 2)

If we know that \vec u =(2,1,1,2) and \vec a=(4,-4,2,-2), then we get that vector component of \vec u parallel to \vec a is:

\vec u_{\parallel\,\vec a} = \left[\frac{(2)\cdot (4)+(1)\cdot (-4)+(1)\cdot (2)+(2)\cdot (-2)}{4^{2}+(-4)^{2}+2^{2}+(-2)^{2}} \right]\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\frac{1}{20}\cdot (4,-4,2,-2)

\vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right)

Lastly, we find the vector component of \vec u orthogonal to \vec a by applying this vector sum identity:

\vec  u_{\perp\,\vec a} = \vec u - \vec u_{\parallel\,\vec a} (Eq. 3)

If we get that \vec u =(2,1,1,2) and \vec u_{\parallel\,\vec a} =\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10} \right), the vector component of \vec u is:

\vec u_{\perp\,\vec a} = (2,1,1,2)-\left(\frac{1}{5},-\frac{1}{5},\frac{1}{10},-\frac{1}{10}    \right)

\vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right)

The component of \vec u orthogonal to \vec a is \vec u_{\perp\,\vec a} = \left(\frac{9}{5},\frac{6}{5},\frac{9}{10},\frac{21}{10}\right).

4 0
3 years ago
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