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3 years ago

How much should 5x³+3x²-2x+1 be increased to get 6x² +7

Mathematics

1 answer:

Mnenie [13.5K]3 years ago

5 0

Answer:

-5x³ + 3x³ + 2x + 6

Step-by-step explanation:

if that is truly the complete problem description, then the only possible answer I can think of, that makes any sense :

what do I need to add to 5x³+3x²-2x+1 to get 6x²+7 ?

the answer is

6x²+7 - (5x³+3x²-2x+1) = 6x²+7-5x³-3x²+2x-1 =

= -5x³ + 3x³ + 2x + 6

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Answer:

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Step-by-step explanation:

Data

Brand A amount of 20 pencils cost $6

Brand B amount of 40 pencils cost $8.5

Process

1.- Calculate the unit price of each brand of pencils dividing the cost by the number of pencils

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3 years ago

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

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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

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3 years ago

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3 years ago

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The explanation is shown below:

1. To solve this problem you must apply the proccedure shown below:

2. When you clear the variable x from the first equation, and subtitute it into the second equation, you obtain:

<span>3x−2y=10
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3. When you subsitute y=-2 into the first equation and clear the x, you have:

x=2
</span>

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How much should 5x³+3x²-2x+1 be increased to get 6x² +7​

How much should 5x³+3x²-2x+1 be increased to get 6x² +7