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Alecsey [184]
3 years ago
15

Given that ABD = 76°, which equation could be used to solve problems involving the relationships between DBC and 2ABC?

Mathematics
1 answer:
Sav [38]3 years ago
5 0

It’s B I think. I might be wrong.

You might be interested in
9 in.
dexar [7]

Answer:

a, a, d

Step-by-step explanation:

44

The area (A) of a triangle is calculated as

A = \frac{1}{2} bh ( b is the base and h the perpendicular height )

Here b = 10 and h = 9 , then

A = \frac{1}{2} × 10 × 9 = 5 × 9 = 45 in² → a

45

The area (A) of a parallelogram is

A = bh ( b is the base and h the perpendicular height )

Here b = 2 and h = 4 , then

A = 2 × 4 = 8 m² → a

46

A = bh ( with b = 4 and h = 10 )

A = 4 × 10 = 40 m² → d

6 0
3 years ago
Does anyone know how to get the days 
nadya68 [22]
System of equations:
y = 6x + 104
y = 10x + 60
 
Method : <span>Substitution
Since the both equation is equal to y, you can substitute one equation for the other equation's y:
</span>
y = 6x + 104

10x + 60 = 6x + 104
-6x   -60  =-6x  -60
-------------------------------
4x = 44
----  -----
 4      4

x = 11

Answer : It would take 11 days.



4 0
3 years ago
Read 2 more answers
The Hydro water department has a monthly service charge of $15.50 and a volume charge of $1.73 for every 100 cubic feet of water
iren [92.7K]
It’s c because it’s a increase in their water bill $5.19
4 0
4 years ago
How do you write 9.6004 × 10^15 in standard form?
Naddik [55]

Answer:

9600400000000000 is in standard form

6 0
3 years ago
35. In a simple random sample of 25 high school students, the sample mean of the SAT scores was 1450, and the sample variance wa
Tatiana [17]

Answer:

1450-2.0\frac{30}{\sqrt{25}}=1450-12    

1450+2.0\frac{30}{\sqrt{25}}=1450+12    

And the best option would be:

c. 1450 +/- 12

Step-by-step explanation:

Information provided

\bar X=1450 represent the sample mean for the SAT scores

\mu population mean (variable of interest)

s^2 = 900 represent the sample variance given

n=25 represent the sample size  

Solution

The confidence interval for the true mean is given by :

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The sample deviation would be s=\sqrt{900}= 30

The degrees of freedom are given by:

df=n-1=2-25=24

The Confidence is 0.954 or 95.4%, the value of \alpha=0.046 and \alpha/2 =0.023, assuming that we can use the normal distribution in order to find the quantile the critical value would be z_{\alpha/2} \approx 2.0

The confidence interval would be

1450-2.0\frac{30}{\sqrt{25}}=1450-12    

1450+2.0\frac{30}{\sqrt{25}}=1450+12    

And the best option would be:

c. 1450 +/- 12

3 0
4 years ago
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