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3 years ago

PLEASE HELP! ONE MULTIPLE CHOICE QUESTION!!!!!!!!!!!!

Mathematics

1 answer:

olga55 [171]3 years ago

7 0

We see that this function has a horizontal asymptote that crosses the y-axis at 2. This means that we can eliminate choices A and C.

From there, we can plug in values for points on the graph into the equation to see if they are true.
f(x)=6(1/3)^x+2
f(1)=6(1/3)^1+2
f(1)=2+2
=4 (Gives us the other coordinate)

f(x)=4(1/3)^x+2
f(1)=4(1/3)^1+2
=(4/3)+2
=3.33 (Does not give us the other coordinate)

Our correct answer choice is B

:)

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Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

algol [13]

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

e) $p_v =P(t_{118}$

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis

5 0

3 years ago

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Vitek1552 [10]

The value of this equation is: 4
Answer: 4

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3 years ago

Find the taylor polynomial t3(x) for the function f centered at the number

Vlad [161]

$e^{-3x}=\displaystyle\sum_{n=0}^\infty\frac{(-3x)^n}{n!}=1-3x+9x^2+\cdots$
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$e^{-3x}\sin2x=\left(1-3x+9x^2+\cdots\right)\left(2x-\dfrac{4x^3}3+\cdots\right)$
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Yuki888 [10]

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3 years ago

Use the picture below to find the length of x. Round your answer to the nearest hundredth.

valentinak56 [21]

Answer: 6.71

=======================================

Work Shown:

The longest horizontal portion of length 6 breaks up into two equal pieces of length 3 each. Focus on the smaller right triangle on the right hand side. This right triangle has legs of 3 and 6. The hypotenuse is x.

Use the pythagorean theorem with a = 3, b = 6, c = x to find the value of x

a^2 + b^2 = c^2

3^2 + 6^2 = x^2

9 + 36 = x^2

45 = x^2

x^2 = 45

x = sqrt(45)

x = 6.7082039

x = 6.71

4 0

3 years ago