Ask question

Ask question

All categories

2 years ago

5

In ∆GHI and ∆JKL, GH = JK, HI = KL, GI = 9’, m∠H = 45°, and m∠K = 65°. Which of the following is not possible for JL: 5’, 9’ or

11’?

2 answers:

AnnZ [28]2 years ago

8 0

Answer:

11'

Step-by-step explanation:

Aliun [14]2 years ago

4 0

Answer:

hello

Step-by-step explanation:

You might be interested in

A line segment is partitioned with a ratio 2:5, how many total parts has the segment been split into

schepotkina [342]

Answer:

7 parts

Step-by-step explanation:

To get the answer to this, we only simply to know the number of times in which the division was made.

This means that the total ratio of the lines is 2 +5 = 7. Meaning we have 7 different divisions.

The first division carries 2/7 while the second division carrues 5/7

7 0

3 years ago

Name two solutions for each inequality.

ahrayia [7]

So,

#1: $n \geq 3 \frac{11}{16} + 4 \frac{1}{2}$

Convert to like improper fractions.
$n \geq \frac{59}{16} + \frac{72}{16}$

Add.
$n \geq \frac{131}{16}\ or\ 8 \frac{3}{16}$

So, one solution could be $8 \frac{3}{16}$ .

Another solution could by 9. There is also 10, 11, 12, etc., and all numbers in between.

#2: $k \ \textless \ 6 \frac{2}{5} * 15$

Convert into improper fraction form.
$k \ \textless \ \frac{32}{5} * 15$

Multiply.
$\frac{(2^5)(3)(5)}{5}$

Cross-cancel, and we have our final result.
$(2^5)(3) = 96$
k < 96

96 is not a solution.

95 is a solution.

So is 94, 93, 92, etc, and all numbers in between.

6 0

3 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

2 years ago

Jacob is cutting yarn for a craft project. He needs pieces of yarn that are 12 inches long, but he only has a ruler marked with

lara [203]

Answer:

30.48 centimeters

Step-by-step explanation:

We know that:

2.54 cm = 1 inch

To find how many centimeters are in 12 inches, you should:

2.54 x 12 = 30.48

So Jacob’s pieces of yarn should be 30.48 centimeters long.

Hope this helps!

5 0

3 years ago

Read 2 more answers

4/5 ÷ 2/3 i'm small brain so I have no clue ;w;

Ad libitum [116K]

Answer:

Exact Form : 6/5

Decimal Form : 1.2

Mixed Nuumber Form ; 1 1/5

3 0

2 years ago

Read 2 more answers