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KATRIN_1 [288]
3 years ago
14

If f(x)=4x-7 and g(x)=2x+4 evalvate f(x)+g(x) for x=-3

Mathematics
2 answers:
galben [10]3 years ago
7 0

Answer:

-21

Step-by-step explanation:

We are told to find f(x) + g(x) for x= -3. Therefore, we must evaluate f(-3) and g(-3), then add them together.

First, evaluate f(-3).

f(x)=4x-7

To find f(-3), we need to substitute -3 in for x.

f(-3)= 4(-3)-7

Solve according to PEMDAS: Parentheses, Exponents, Multiplication, Division, Addition, Subtraction First, multiply 4 and -3.

f(-3)= -12-7

Next, subtract 7 from -12

f(-3)= -19

Next, find g(-3).

g(x)=2x+4

To find g(-3), substitute -3 in for x.

g(-3)= 2(-3)+4

Solve according to PEMDAS. First, multiply 2 and -3.

g(-3)= -6+4

Next, add -6 and 4

g(-3)= -2

Now, we can add f(-3) and g(-3) together.

f(-3) + g(-3)

f(-3)= -19

g(-3)= -2

-19 + -2

Add

-21

Setler79 [48]3 years ago
7 0

Answer:

-21

Step-by-step explanation:

Adding f and g together, we get  (f + g)(x) = 4x + 2x -7 + 4, or

                                                                      = 6x - 3

Now replace x with -3.  We get:

(f + g)(-3) = 6(-3) - 3 = -21

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How do you convert a fraction to a percent
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Some researchers have conjectured that stem-pitting disease in peach-tree seedlings might be controlled with weed and soil treat
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Answer:

Part A

b. 14.6 ± 7.38

Part B

b. 3.43

Part C

a. P-value < 0.01

Part D

b. There is sufficient evidence to reject the null hypothesis

Step-by-step explanation:

Part A

The given data are;

The number of seedlings in the field = 20

The number of seedlings selected to receive herbicide A = 10

The number of seedlings selected to receive herbicide B = 10

The height in centimeters of seedlings treated with Herbicide A, \overline x _1 = 94.5 cm

The standard deviation, s₁ = 10 cm

The height in centimeters of seedlings treated with Herbicide B, \overline x _2 = 109.1 cm

The standard deviation, s₂ = 9 cm

The 90% confidence interval for μ₂ - μ₁, is given as follows;

\left (\bar{x}_{2}- \bar{x}_{1}  \right )\pm t_{\alpha /2}\sqrt{\dfrac{s_{1}^{2}}{n_{1}}+\dfrac{s_{2}^{2}}{n_{2}}}

The critical-t at 95% and n₁ + n₂ - 2 degrees of freedom is given as follows;

The degrees of freedom, df = n₁ + n₂ - 2 = 10 + 10 - 2 = 18

α = 100% - 90% = 10%

∴ For two tailed test, we have, α/2 = 10%/2 = 5% = 0.05

t_{(0.025, \, 18)} = 1.734

C.I. = \left (109.1- 94.5 \right )\pm 1.734 \times \sqrt{\dfrac{10^{2}}{10}+\dfrac{9^{2}}{10}}

C.I. ≈ 14.6 ± 7.37714603353

The 90% C.I. ≈ 14.6 ± 7.38

b. 14.6 ± 7.38

Part B

With the hypotheses are given as follows;

H₀; μ₂ - μ₁ = 0

Hₐ; μ₂ - μ₁ ≠ 0

The two sample t-statistic is given as follows;

t=\dfrac{(\bar{x}_{2}-\bar{x}_{1})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}+\dfrac{s _{2}^{2}}{n_{2}}}}

t-statistic=\dfrac{(109.1-94.5)}{\sqrt{\dfrac{10^{2} }{10}+\dfrac{9^{2}}{10}}} \approx 3.43173361147

The two sample t-statistic ≈ 3.43

b. 3.43

Part C

From the t-table, the p-value, we have, the p-value < 0.01

a. P-value < 0.01

Part D

Given that a significance level of 0.05 level is used and the p-value of 0.01 is less than the significance level, there is enough statistical evidence to reject the null hypothesis

b. There is sufficient evidence to reject the null hypothesis.

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