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olga nikolaevna [1]
4 years ago
12

( 2 + 3 ) + ( 4 + x )

Mathematics
1 answer:
zzz [600]4 years ago
3 0
2+3=5 4+x=4x 5+4x=9x
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SOVA2 [1]
The answer is 2

RISE OVER RUN! :)
5 0
3 years ago
I’ve been stuck on this for a while now and i can’t get passed it can someone please help me out please please help
KonstantinChe [14]

Answer:

(D.) $5.35

Step-by-step explanation:

1.25+0.35=1.60

1.60+3.75=5.35

4 0
3 years ago
The amount of time it takes a bat to eat a frog was recorded for each bat in a random sample of 12 bats. The resulting sample me
spin [16.1K]

Answer: a. CI for the mean: 17.327 < μ < 26.473

b. CI for variance: 29.7532 ≤ \sigma^{2} ≤ 170.9093

Step-by-step explanation:

a. To construct a 95% confidence interval for the mean:

The given data are:

mean = 21.9

s = 7.7

n = 12

df = 12 - 1 = 11

1 - α = 0.05

\frac{\alpha}{2} = 0.025

t-score = t_{0.025,11} = 2.2001

Note: since the sample population is less than 30, it is used a t-score.

The formula for interval:

mean ± t.\frac{s}{\sqrt{n} }

Substituing values:

21.9 ± 2.200.\frac{7.7}{\sqrt{12} }

21.9 ± 4.573

The interval is: 17.327 < μ < 26.473

b. A 95% confidence interval for the variance:

The given values are:

s^{2} = 7.7^{2}

s^{2} = 59.29

α = 0.05

\frac{\alpha}{2} = 0.025

1-\frac{\alpha}{2} = 0.975

\chi^{2}_{0.025,11} = 21.92

\chi^{2}_{0.975,11} = 3.816

Note: To find the values for \chi^{2}_{\alpha/2,n-1} and \chi^{2}_{1-\alpha/2,n-1}, look for them at the chi-square table

The formula to calculate interval:

(\frac{(n-1).s^{2}}{\chi^{2}_{\alpha/2,n-1}} , \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2,n-1}})

are the lower and upper limits, respectively.

Substituing values:

(\frac{11.59.29}{21.92} , \frac{11.59.29}{3.816})

(29.7532, 170.9093)

The interval for variance is: 29.7532 ≤ \sigma^{2} ≤ 170.9093

6 0
3 years ago
What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?
PilotLPTM [1.2K]

ANSWER


The orthocenter is (-4,8)


<u>EXPLANATION</u>


The orthocenter is  the point of intersection of all the altitudes of a triangle.

If it is a right angle triangle then the orthocenter is the point at which the <em>right angle is made.</em>

If it is not a right angle triangle, then you have to find the equations of any two of the altitudes and solve simultaneously.

So you must first check to see if it is a right angle triangle that way you do not have to do bunch work.


The given triangle has vertices J(-4,-1), (-4,8), and L(2,8).


Slope_{JK}=\frac{8--1}{-4--4}


Slope_{JK}=\frac{8+1}{-4+4}


Slope_{JK}=\frac{9}{0}


Since the denominator is zero, the slope is undefined.

If the slope of a straight line is undefined, then that line is <em>parallel to the y-axis</em>. In other words, it is a <em>vertical line</em>.


Slope_{JL}=\frac{8--1}{2--4}


Slope_{JL}=\frac{8+1}{2+4}


Slope_{JL}=\frac{9}{6}


Slope_{JL}=\frac{3}{2}



Slope_{KL}=\frac{8-8}{2--4}


Slope_{KL}=\frac{0}{2+4}


Slope_{KL}=\frac{0}{6}


Slope_{KL}=0


if a straight line has a slope of zero, that line is parallel to the <em>x-axis</em> or it is <em>horizontal.</em>


From the slopes we can see that side KL of the triangle is horizontal and side JK is vertical. This two sides will meet at right angles at K.

See graph.

Therefore the coordinates of K, (-4,8) is the orthocenter of the given triangle.







3 0
3 years ago
Please help me solve this question thank you​
GarryVolchara [31]

Answer:

Step-by-step explanation:

7 0
4 years ago
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