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Artist 52 [7]
3 years ago
7

F(x)=2x^2-x-6 g(x)=4-x (f+g)(x)

Mathematics
1 answer:
Semenov [28]3 years ago
8 0

\large  \boxed{(f + g)(x) = f(x) + g(x)}

The property above is distribution property where we distribute x-term in the function.

Substitute both f(x) and g(x) in.

\large{ \begin{cases} f(x) = 2 {x}^{2}   -  x - 6 \\ g(x) =  4 - x \end{cases}}  \\  \large{f(x) + g(x) = (2 {x}^{2}  - x - 6) + (4 - x)} \\  \large{f(x) + g(x) = 2 {x}^{2}  - x - 6 + 4 - x}

Évaluate/Combine like terms.

\large{f(x)  + g(x) = 2 {x}^{2}  - 2x - 2}

The function can be factored so there are two answers. (Both of them work as one of them is factored form while the other one is not.)

\large{(f + g)(x) = 2 {x}^{2}  -2x -2}

<u>Alternative</u>

\large{(f + g)(x) = 2({x}^{2}  -x - 1)}

<u>Answer</u>

  • (f+g)(x) = 2x²-2x-2
  • (f+g)(x) = 2(x²-x-1)

Both answers work. The second answer is in factored form.

Let me know if you have any doubts!

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almond37 [142]

Answer:

You need to find the x and y intercepts so...

The x intercept is (18,0) since x=18

The y intercept is (0,12) since y=12

Step-by-step explanation:

Cover the y when you are finding the y intercept

Cover the x when you are finding the x intercept

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3 years ago
Two 5-year girls, Alyse and Jocelyn, have been training to run a 1-mile race. Alyse's 1 mile time A is approximately Normally di
tatyana61 [14]

Answer:

1.7 × 10⁻⁴

Step-by-step explanation:

The question relates to a two sample z-test for the comparison between the means of the two samples  

The null hypothesis is H₀:  μ₁ ≤ μ₂

The alternative hypothesis is Hₐ: μ₁ > μ₂

z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}

Where;

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\bar {x}_2 = 12

σ₁ = 2.5

σ₂ = 1.5

We set our α level at 0.05

Therefore, our critical z = ± 1.96

For n₁ = n₂ = 23, we have;

z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^{2} }{23}-\dfrac{1.5^{2}}{23}}} = 3.5969

We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore;

The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.

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Step-by-step explanation:

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