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2 years ago

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A company makes traffic signs.One of their signs can be modeled by an equilateral triangle with a perimeter of 144 inches. The c

ompany makes a similar sign with perimeter that is 1.25 times the one shown. What is the height in inches of the larger sign?

1 answer:

Mars2501 [29]2 years ago

6 0

Answer:

The perimeter of the larger sign : 144*1,25= 180 inches

Side of the triangle= 180/3= 60 inches

square of the height= 60^(2) - 30^(2)= 2700 ( Pythagoras' theorem)

height= square root ( 2700)= 51,96

Step-by-step explanation:

You might be interested in

A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fis

maks197457 [2]

Answer:

p = 0.293

Conclusion:

"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."

Step-by-step explanation:

------------R/trout--O/trout---bass--catfish---total

GVL - - - 105 - - - 27 - - - - - 35--- 24 - - - 191

EL - - - - 135------- 48-------- 67 - - 43 - - - 293

Total - - 240 - - - 75 - - - - - 102--- 67 - - - 484

Note : GVL = GREEN VALLEY LAKE ; EL = ECHO LAKE.

H0: The distribution of fish caught in green valley is the same as that caught in Echo lake.

H1: The distribution of fish caught in Green Valley lake is different from that caught in Echo valley lake.

Using the chi-squared test statistic calculator ;

χ2 = 3.7269

p-value = 0.2925.

p-value = 0.293

Conclusion:

"There is insufficient evidence to state that the distribution of fish in Green Valley Lake is not the same as the distribution of fish in Echo Lake."

5 0

2 years ago

Lenovo uses the zx-81 chip in some of its laptop computers. the prices for the chip during the last 12 months were as follows:

Stella [2.4K]

Given the table below of the prices for the Lenovo zx-81 chip during the last 12 months

$\begin{tabular}
{|c|c|c|c|}
Month&Price per Chip&Month&Price per Chip\\[1ex]
January&\$1.90&July&\$1.80\\
February&\$1.61&August&\$1.83\\
March&\$1.60&September&\$1.60\\
April&\$1.85&October&\$1.57\\
May&\$1.90&November&\$1.62\\
June&\$1.95&December&\$1.75
\end{tabular}$

The forcast for a period $F_{t+1}$ is given by the formular

$F_{t+1}=\alpha A_t+(1-\alpha)F_t$

where $A_t$ is the actual value for the preceding period and $F_t$ is the forcast for the preceding period.

Part 1A:
Given <span>α = 0.1 and the initial forecast for october of $1.83, the actual value for october is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha A_{10}+(1-\alpha)F_{10} \\ \\ =0.1(1.57)+(1-0.1)(1.83) \\ \\ =0.157+0.9(1.83)=0.157+1.647 \\ \\ =1.804$

Therefore, the foreast for period 11 is $1.80

Part 1B:

</span>Given <span>α = 0.1 and the forecast for november of $1.80, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.1(1.62)+(1-0.1)(1.80) \\ \\ 
=0.162+0.9(1.80)=0.162+1.62 \\ \\ =1.782$

Therefore, the foreast for period 12 is $1.78</span>

Part 2A:

Given <span>α = 0.3 and the initial forecast for october of $1.76, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.3(1.57)+(1-0.3)(1.76) \\ \\ 
=0.471+0.7(1.76)=0.471+1.232 \\ \\ =1.703$

Therefore, the foreast for period 11 is $1.70

</span>
<span><span>Part 2B:

</span>Given <span>α = 0.3 and the forecast for November of $1.70, the actual value for november is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.3(1.62)+(1-0.3)(1.70) \\ \\ 
=0.486+0.7(1.70)=0.486+1.19 \\ \\ =1.676$

Therefore, the foreast for period 12 is $1.68

</span></span>
<span>Part 3A:

Given <span>α = 0.5 and the initial forecast for october of $1.72, the actual value for October is $1.57.

Thus, the forecast for period 11 is given by:

$F_{11}=\alpha
 A_{10}+(1-\alpha)F_{10} \\ \\ =0.5(1.57)+(1-0.5)(1.72) \\ \\ 
=0.785+0.5(1.72)=0.785+0.86 \\ \\ =1.645$

Therefore, the forecast for period 11 is $1.65

</span>
<span><span>Part 3B:

</span>Given <span>α = 0.5 and the forecast for November of $1.65, the actual value for November is $1.62

Thus, the forecast for period 12 is given by:

$F_{12}=\alpha
 A_{11}+(1-\alpha)F_{11} \\ \\ =0.5(1.62)+(1-0.5)(1.65) \\ \\ 
=0.81+0.5(1.65)=0.81+0.825 \\ \\ =1.635$

Therefore, the forecast for period 12 is $1.64

Part 4:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span></span></span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.83, $1.80, $1.78

Thus, the mean absolute deviation is given by:

$\frac{|1.57-1.83|+|1.62-1.80|+|1.75-1.78|}{3} = \frac{|-0.26|+|-0.18|+|-0.03|}{3} \\ \\ = \frac{0.26+0.18+0.03}{3} = \frac{0.47}{3} \approx0.16$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.1 of October, November and December is given by: 0.157

</span><span><span>Part 5:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.3, we obtained that the forcasted values of october, november and december are: $1.76, $1.70, $1.68

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.76|+|1.62-1.70|+|1.75-1.68|}{3} = 
\frac{|-0.17|+|-0.08|+|-0.07|}{3} \\ \\ = \frac{0.17+0.08+0.07}{3} = 
\frac{0.32}{3} \approx0.107$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.3 of October, November and December is given by: 0.107

</span></span>
<span><span>Part 6:

The mean absolute deviation of a forecast is given by the summation of the absolute values of the actual values minus the forecasted values all divided by the number of items.

Thus, given that the actual values of october, november and december are: $1.57, $1.62, $1.75

using </span><span>α = 0.5, we obtained that the forcasted values of october, november and december are: $1.72, $1.65, $1.64

Thus, the mean absolute deviation is given by:

$
 \frac{|1.57-1.72|+|1.62-1.65|+|1.75-1.64|}{3} = 
\frac{|-0.15|+|-0.03|+|0.11|}{3} \\ \\ = \frac{0.15+0.03+0.11}{3} = 
\frac{29}{3} \approx0.097$

Therefore, the mean absolute deviation </span><span>using exponential smoothing where α = 0.5 of October, November and December is given by: 0.097</span></span>

5 0

1 year ago

What is the value of b in the equation (y Superscript b Baseline) Superscript 4 Baseline = StartFraction 1 Over y Superscript 24

shusha [124]

The value of b is -6.

Explanation:

The expression is $\left(y^{b}\right)^{4}=\frac{1}{y^{24}}$

To determine the value of b, we shall solve the expression.

Applying exponent rule, $\left(a^{b}\right)^{c}=a^{b c}$ , we get,

$y^{4b}=\frac{1}{y^{24}}$

Applying exponent rule, $\frac{1}{a^{b}}=a^{-b}$ , we have,

$y^{4b}=y^{-24}$

The expression is of the form, $a^{f(x)}=a^{g(x)}$ then $f(x)=g(x)$

Applying this rule, we get,

$4b=-24$

Dividing both sides by 4, we have,

$b=-6$

Hence, the value of b is -6.

4 0

2 years ago

Read 2 more answers

A case in riverhead, new york, nine different crime victims listened to voice recordings of five different men. all nine victims

Alexus [3.1K]

the complete question in the attached figure

1) Find the probability that all nine victims would select the same person. You should enter your answer as a decimal, not as a percent. Do not round, as your answer should be a terminating decimal.

P=(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)=0.000000512

2) what is the probability that the first victim picks SOMEONE? ANYONE?

P=(1/5)=0.20 ---------------- > 20%

3) Now let’s look at victim #2. What is the probability that they pick the same person as victim #1?

P=(1/5)*(1/5)=0.04 --------------> 4%

4) Now let’s look at victim #3. What is the probability that they pick the same person as victim #1?

P=(1/5)*(1/5)*(1/5)=0.008 --------------> 0.8%

5) Now let's look at victim #9. What is the probability that they pick the same person as victim #1?

P=(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)*(1/5)=0.000000512

6) Once you find all those probabilities, what rule would apply?

the probability is given by (1/5)^n

n=number of victims that pick the same person---------> in this problem = 9

P=(1/5)^9= 0.000000512

5 0

2 years ago

Which is greater 9000cm or 2 km?

satela [25.4K]

Convert 2 km into cm

1 km = 100,000 cm

2 km = 200,000 cm

therefore 9,000 cm is greater

6 0

1 year ago