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vivado [14]
3 years ago
9

Triangle EFG is similar to triangle HIJ. Find the measure of side JH. Round your

Mathematics
1 answer:
MAVERICK [17]3 years ago
8 0

Answer:

JH = 97.9

Step-by-step explanation:

25/12 = x/47

12x = 1175

x = 97.917

You might be interested in
Spotlight Dance increased the number of dancers on its competition team. Last
Nana76 [90]

Answer:

80%

Step-by-step explanation:

To find percentage increase, we use the following formula:

Percentage Increase = \frac{New-Old}{Old}*100

Here

New means the latest value (which is 18 dancers)

Old means the previous value (which is 10 dancers)

The Multiplication by 100 is to convert the decimal answer to percentage.

So, we have:

New = 18

Old = 10

Now we substitute and find:

\frac{New-Old}{Old}*100\\\frac{18-10}{10}*100\\\frac{8}{10}*100\\80

So, there was an increase of 80%

3 0
3 years ago
Solve and state any extraneous solutions
Mice21 [21]

\frac{3x}{x - 1}  + 2 =  \frac{3}{x - 1}

\frac{3x}{x - 1} +  \frac{2(x - 1) }{ x - 1}   =  \frac{3}{x - 1}

\frac{3x}{x - 1}  +  \frac{2x - 2}{ x - 1}  =  \frac{3}{x - 1}

\frac{3x + 2x - 2}{ x - 1}  =  \frac{3}{x - 1}

3x + 2x - 2 = 3

5x - 2 = 3

5x = 3 + 2

5x = 5

x =  \frac{5}{5}

x = 1

3 0
3 years ago
The area of the shaded segment is 100cm^2. Calculate the value of r.
Reil [10]
Hello, 

The formula for finding the area of a circular region is: A=  \frac{ \alpha *r^{2} }{2}

then:
A_{1} = \frac{80*r^{2} }{2}

With the two radius it is formed an isosceles triangle, so, we must obtain its area, but first we obtain the height and the base.

cos(40)= \frac{h}{r}  \\  \\ h= r*cos(40)\\ \\ \\ sen(40)= \frac{b}{r} \\ \\ b=r*sen(40)

Now we can find its area:
A_{2}=2* \frac{b*h}{2}  \\  \\ A_{2}= [r*sen(40)][r*cos(40)]\\  \\A_{2}= r^{2}*sen(40)*cos(40)

The subtraction of the two areas is 100cm^2, then:

A_{1}-A_{2}=100cm^{2} \\ (40*r^{2})-(r^{2}*sen(40)*cos(40) )=100cm^{2} \\ 39.51r^{2}=100cm^{2} \\ r^{2}=2.53cm^{2} \\ r=1.59cm

Answer: r= 1.59cm
7 0
3 years ago
Read 2 more answers
Is the expression 3^2 ^3 equivalent to 3^3 ^2<br> explain.
Pavlova-9 [17]

Answer:

no

Step-by-step explanation:

because they have 2 different answers

4 0
3 years ago
Read 2 more answers
A 95% confidence interval was computed using a sample of 16 lithium batteries, which had a sample mean life of 645 hours. The co
polet [3.4K]

Answer:

Step-by-step explanation:

Hello!

The mean life of 16 lithium batteries was estimated with a 95% CI:

(628.5, 661.5) hours

Assuming that the variable "X: Duration time (life) of a lithium battery(hours)" has a normal distribution and the statistic used to estimate the population mean was s Student's t, the formula for the interval is:

[X[bar]±t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }]

The amplitude of the interval is calculated as:

a= Upper bond - Lower bond

a= [X[bar]+t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] -[X[bar]-t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }]

and the semiamplitude (d) is half the amplitude

d=(Upper bond - Lower bond)/2

d=([X[bar]+t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] -[X[bar]-t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }] )/2

d= t_{n-1;1-\alpha /2}* \frac{S}{\sqrt{n} }

The sample mean marks where the center of the calculated interval will be. The terms of the formula that affect the width or amplitude of the interval is the value of the statistic, the sample standard deviation and the sample size.

Using the semiamplitude of the interval I'll analyze each one of the posibilities to see wich one will result in an increase of its amplitude.

Original interval:

Amplitude: a= 661.5 - 628.5= 33

semiamplitude d=a/2= 33/2= 16.5

1) Having a sample with a larger standard deviation.

The standard deviation has a direct relationship with the semiamplitude of the interval, if you increase the standard deviation, it will increase the semiamplitude of the CI

↑d= t_{n-1;1-\alpha /2} * ↑S/√n

2) Using a 99% confidence level instead of 95%.

d= t_{n_1;1-\alpha /2} * S/√n

Increasing the confidence level increases the value of t you will use for the interval and therefore increases the semiamplitude:

95% ⇒ t_{15;0.975}= 2.131

99% ⇒ t_{15;0.995}= 2.947

The confidence level and the semiamplitude have a direct relationship:

↑d= ↑t_{n_1;1-\alpha /2} * S/√n

3) Removing an outlier from the data.

Removing one outlier has two different effects:

1) the sample size is reduced in one (from 16 batteries to 15 batteries)

2) especially if the outlier is far away from the rest of the sample, the standard deviation will decrease when you take it out.

In this particular case, the modification of the standard deviation will have a higher impact in the semiamplitude of the interval than the modification of the sample size (just one unit change is negligible)

↓d= t_{n_1;1-\alpha /2} * ↓S/√n

Since the standard deviation and the semiamplitude have a direct relationship, decreasing S will cause d to decrease.

4) Using a 90% confidence level instead of 95%.

↓d= ↓t_{n_1;1-\alpha /2} * S/√n

Using a lower confidence level will decrease the value of t used to calculate the interval and thus decrease the semiamplitude.

5) Testing 10 batteries instead of 16. and 6) Testing 24 batteries instead of 16.

The sample size has an indirect relationship with the semiamplitude if the interval, meaning that if you increase n, the semiamplitude will decrease but if you decrease n then the semiamplitude will increase:

From 16 batteries to 10 batteries: ↑d= t_{n_1;1-\alpha /2} * S/√↓n

From 16 batteries to 24 batteries: ↓d= t_{n_1;1-\alpha /2} * S/√↑n

I hope this helps!

4 0
3 years ago
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