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PIT_PIT [208]
2 years ago
8

6.) Find the measure of a. 115°

Mathematics
1 answer:
-Dominant- [34]2 years ago
4 0

Answer:

65 degrees :)

Step-by-step explanation:

If you have 115 degrees, and you know triangles are always 180 degrees in total, then you have to subtract 115 from 180 to get your answer for what the measures are. 180-115=65

So  it's 65 degrees.

I think this is what you want to know :)

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The product of two consecutive odd integers is 782. What are the integers?
Alex Ar [27]

Call the smaller of the two odds = n

Call the next number in the sequence = n + 2

n*(n +2) = 782           Remove the brackets.

n^2 + 2n = 782          Subract 782 from both sides.

n^2 + 2n - 782 = 0    We are going to have to factor this.

Discussion

This problem can't be done the way it is written. The product of an odd integer with another odd integer is and odd integer. There are no exceptions to this. So you need to give a number that has two factors very near it's square root for this question to work.

For example, you could use 783, (which factors) instead of 782 .

Solve

n^2 + 2n - 783 = 0

(n + 29)(x - 27) = 0

<u>Solution One</u>

n - 27 = 0

n = 27

The two odd consecutive integers are 27 and 29.

<u>Solution Two</u>

n + 29 = 0

n = - 29

The two solution integers are -29 and - 27 Notice that - 29 is smaller than - 27.

7 0
3 years ago
The median is 28; 16,24,x,48
Nookie1986 [14]
x = 32 

To find the media, you take the middle number. If there is no discrete middle number, take the sum of the two closest to the middle and divide by 2. 

(24 + x)/2 = 28

Multiply by 2 on both sides to get:

24 + x = 56

Subtract 24 from both sides to isolate the variable:

x = 56 - 24
x = 32
3 0
3 years ago
What the nearest hundred thousand for 767074
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7 is  digit for  hundred thousands and 6  is for ten thousands so the answer is 800,000
5 0
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Diana invested $3000 in a savings account for 3 years. She earned $450 in interest over that time period. What interest rate did
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\bf \qquad \textit{Simple Interest Earned}\\\\&#10;I = Prt\qquad &#10;\begin{cases}&#10;I=\textit{interest earned}\to &\$450\\&#10;P=\textit{original amount deposited}\to& \$3000\\&#10;r=rate\to r\%\to \frac{r}{100}\\&#10;t=years\to &3&#10;\end{cases}&#10;\\\\\\&#10;450=3000\cdot r\cdot 3\implies \cfrac{450}{3\cdot 3000}=r\implies \cfrac{1}{20}=r&#10;\\\\\\&#10;r\%\implies \cfrac{1}{20}\cdot 100\implies r=\stackrel{\%}{5}
5 0
3 years ago
Read 2 more answers
An insurance company selected samples of clients under 18 years of age and over 18 and recorded the number of accidents they had
Stolb23 [73]

Answer:

Q1 z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the that means have no difference

Q2  CI 95 %  =  (  0,056 ;  0,164 )

Step-by-step explanation:

Sample information for people under 18

n₁  =  500

x₁ =  180

p₁  =  180/ 500    p₁  =  0,36    then  q₁  =  1 -  p₁     q₁ =  0,64

Sample information for people over 18

n₂  =  600

x₂  =  150

p₂  =  150 / 600   p₂ =  0,25   then   q₂  =  1 - p₂   q₂ =  1 - 0,25   q₂ = 0,75

Hypothesis Test

Null hypothesis                        H₀              p₁  =  p₂

Alternative Hypothesis           Hₐ              p₁  ≠  p₂

The alternative hypothesis indicates that the test is a two-tail test.

We will use the approximation to normal distribution of the binomial distribution according to the sizes of both samples.

Testin at CI =  95 %    significance level is  α = 5 %   α  =  0,05  and

α/ 2  =  0,025   z (c) for that α  is from z-table:

z(c) = 1,96

To calculate   z(s)

z(s)  =  ( p₁  -   p₂ ) / EED

EED = √(p₁*q₁)n₁  +  (p₂*q₂)/n₂

EED = √( 0,36*0.64)/500  +  (0,25*0,75)/600

EED = √0,00046  +  0,0003125

EED = 0,028

( p₁  -  p₂  )  =  0,36  -  0,25  = 0,11

Then

z(s)  =  0,11 / 0,028

z(s) = 3,93

Comparing  z(s) and  z (c)    z(s) > z(c)

z(s) is in the rejection region for H₀ ; we reject H₀. We can´t support the idea of equals means

Q2  CI  95 %   =  (  p₁  -  p₂  ) ±  z(c) * EED

CI 95%  =  ( 0,11   ±  1,96 * 0,028 )

CI 95%  = (  0,11  ±  0,054 )

CI 95 %  =  (  0,056 ;  0,164 )

8 0
2 years ago
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