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trasher [3.6K]
3 years ago
12

PLEASE HELP ASAP FIND THE LENGTH

Mathematics
1 answer:
Bond [772]3 years ago
5 0
Triangle ABC:
<ABC=90°
<ACB=y

Triangle CDE:
<CDE=90°
<DCE=y

The two triangles have two congruent angles, then they are similars, and theirs sides must be proportionals:

Adjacent angle to y / Hypotenuse

Triangle CDE       Triangle ABC
CD/CE             =       BC/AC

Replacing the values:
x/7=(20-x)/21

Solving for x. Cross multiplication:
(21)(x)=(7)(20-x)
21x=140-7x
21x+7x=140-7x+7x
28x=140
28x/28=140/28
x=5

Length of CD = x = 5

Answer: The length of CD is 5

Answer: 5  
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Anvisha [2.4K]
Three adds to get 5 and also it multiplies to get 24 

Hope i Helped

4 0
3 years ago
What is the algerbraic expression for the following word phrase: the product of 5 more than p and 7?
konstantin123 [22]

Answer:

Step-by-step explanation:

7(p+5)

5 0
3 years ago
A book claims that more hockey players are born in January through March than in October through December. The following data sh
astra-53 [7]

Answer:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

Step-by-step explanation:

We need to conduct a chi square test in order to check the following hypothesis:

H0: There is no difference of birthdates distributed throughout the​ year

H1: There is a difference between birthdates distributed throughout the​ year

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{total}{4}

And replacing we got:

E_{1} =\frac{67+56+30+37}{4}=47.5

And now we can calculate the statistic:

\chi^2 = \frac{(67-47.5)^2}{47.5}+\frac{(56-47.5)^2}{47.5}+\frac{(30-47.5)^2}{47.5}+\frac{(37-47.5)^2}{47.5}=18.295

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=4-1=3

And we can calculate the p value given by:

p_v = P(\chi^2_{3} >18.295)=0.00038

Since the p value is very low we have enough evidence to reject the null hypothesis and we can conclude that the players' birthdates are not uniformly distributed throughout the​ year

3 0
4 years ago
Find the area of the shaded sector of the circle
sweet-ann [11.9K]

Answer:

119.78 square m.

Step-by-step explanation:

d = 28 \: m \implies \: r = 14 \: m \\ central \:  \angle \:( \theta)  \\ = 180 \:degree - 110 \degree \\  = 70 \degree \\  \\ area \: of \: sector  \\ =  \frac{ \theta}{360 \degree}  \times \pi {r}^{2}  \\ \\   =  \frac{70 \degree}{360 \degree}  \times  \frac{22}{7}   \times {14}^{2}  \\  \\  = \frac{7}{36}  \times  \frac{22}{7}   \times 196 \\  \\  = \frac{7}{36}  \times  \frac{22}{1}   \times 28 \\ \\  = \frac{7}{9}  \times  22   \times 7 \\  \\  =  \frac{1078}{9}  \\  \\  = 119.777778 \\  \\  = 119.78 \:  {m}^{2}  \\

3 0
4 years ago
PLease answer!!! 30 points!!!
Agata [3.3K]

Answer:

a

Step-by-step explanation:

They make a right angle. A right angle adds up to 90.

7 0
3 years ago
Read 2 more answers
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