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kari74 [83]
3 years ago
8

Which equation shows the substitution method being used to solv the system of equations?

Mathematics
1 answer:
slava [35]3 years ago
8 0
The x value in the second equation (y+2) is substituted for the x in the first equation in...

Answer A.
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Complete the two-column the proof.
Serga [27]
1. ∠1 is complementary to ∠2. 1. givem

2.  m∠1 + m∠2.  2. Definition of complementary

3.BD bisects ∠ADC. 3. Given

4.  4. Definition of bisect

5. m∠2 = m∠3 5.  

6. m∠1 + m∠3 = 90° 6.  sunsitution

7. 7.
5 0
3 years ago
Read 2 more answers
consider the exponential function f(x)= 1/5(15x) what is the value of the growth factor of the function?
Rus_ich [418]

Answer:

  15

Step-by-step explanation:

The general form of an exponential equation is ...

  f(x) = (initial value)(growth factor)^x

That is, the "growth factor" is the base of the exponent. In your equation ...

  f(x) = (1/5)(15^x)

the growth factor is 15.

3 0
3 years ago
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Best Answer Gets Brainliest Please help asap
FrozenT [24]
<h2>Answer:</h2>

[1] Area of base = 13 × 13 = 169in².

Area of faces = 4 (1/2 × 13 × 8) = 208in².

Surface area = (169 + 208)in² = 377in².

[2] Area of base = 1/2 × 5.2 × 4.5 = 11.7in².

Area of faces = 3 (√3/4 × 5.2²) = 35.1in².

Surface area = (35.1 + 11.7)in² = 46.8in².

[3] Area of base = 7 × 10 = 70in².

Area of faces = 2(1/2 × 7 × 6) + 2(1/2 × 10 × 4.8)

= 98in².

Surface area = (70 + 98)in² = 168in².

7 0
3 years ago
What is the median 18 16 -14 -15 8 18 7 11 17
Dominik [7]

Answer:

11

Step-by-step explanation:

-15, -15, 7, 8, 11, 16, 17, 18, 18

find the middle number after numbers are in order

hope this helps :)

8 0
3 years ago
A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

<u>Step-by-step explanation</u>:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0
3 years ago
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