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Zina [86]
3 years ago
11

The attic floor, ABCD in the model, is a square. The beams that support the roof are the edges of a block (rectangular prism) EF

GHKLMN. E is the middle of AT, F is the middle of BT, G is the middle of CT and H is the middle of DT. All the edges of the pyramid in the model have length 12m.
Calculate the length of EF, one of the horizontal edges of the block.
Mathematics
1 answer:
morpeh [17]3 years ago
8 0

Answer:

EF = 6m.

Step-by-step explanation:

Given that:

1) All the edges of the pyramid in the model have length 12m.

So, AB = 12m

2) E is the middle of AT, F is the middle of BT

So, EF is a line segment connecting the midpoints of ΔATB

So, by applying The Triangle Mid-segment Theorem

EF // AB and EF = 0.5 AB

So, EF = 0.5 AB = 0.5 * 12 = 6m

====================================

The Triangle Mid-segment Theorem:

The line segment connecting the midpoints of any two sides of a triangle has the following properties:

1) The line segment will be parallel to the third side.

2) The length of the line segment will be a half of the length of the third side.

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Lines A and B are parallel to each other. Lines B and E are perpendicular to each other. 
4 0
3 years ago
An investigator thinks that people under the age of forty have vocabularies that are different than those of people over sixty y
Kay [80]

Answer:

t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28  

Now we can calculate the p value with the following probability:

p_v =2*P(t_{60}  

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

Step-by-step explanation:

Information given

\bar X_{1}=14 represent the mean for sample 1 (younger)

\bar X_{2}=20 represent the mean for sample 2 (older)  

s_{1}=5 represent the sample standard deviation for 1  

s_{f}=6 represent the sample standard deviation for 2  

n_{1}=31 sample size for the group 2  

n_{2}=31 sample size for the group 2  

t would represent the statistic

System of hypothesis

We want to test if  that people under the age of forty have vocabularies that are different than those of people over sixty years of age, the system of hypothesis are:

Null hypothesis:\mu_{1}-\mu_{2}=0  

Alternative hypothesis:\mu_{1} - \mu_{2}\neq 0  

The statistic is given by:

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}} (1)  

And the degrees of freedom are given by df=n_1 +n_2 -2=31+31-2=60  

Replacing the info given we got:

t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28  

Now we can calculate the p value with the following probability:

p_v =2*P(t_{60}  

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

6 0
3 years ago
A cookie factory monitored the number of broken cookies per pack yesterday.
trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given  

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20  

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } } \leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

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