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2 years ago

Find the value of the polynomial 5x – 4x^2 + 3 at x = 2 and x = –1

Mathematics

2 answers:

Andrew [12]2 years ago

7 0

Answer:

x=2 => 5×2 - 4×2² + 3 = 10-16+3 = -3

x=-1 => 5×-1 - 4×(-1)² + 3 = -5 - 4 + 3 = -6

Step-by-step explanation:

is in the answer.

remember that squaring a negative number creates a positive result.

aliya0001 [1]2 years ago

6 0

Answer:

Let the polynomial be f(x) = 5x – 4x^2 + 3

Now, for x = 2,

f(2) = 5(2) – 4(2)^2 + 3

=> f(2) = 10 – 16 + 3 = –3

Or, the value of the polynomial 5x – 4x^2 + 3 at x = 2 is -3.

Similarly, for x = –1,

f(–1) = 5(–1) – 4(–1^)2 + 3

=> f(–1) = –5 –4 + 3 = -6

The value of the polynomial 5x – 4x2 + 3 at x = -1 is -6.

Answered by GAUTHMATH

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Answer:

Step-by-step explanation:

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10 months ago

A lathe is set to cut bars of steel into lengths of 6 cm. The lathe is considered to be in perfect adjustment if the average len

Gnom [1K]

Answer:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

$df=n-1=93-1=92$

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

Step-by-step explanation:

Information provided

$\bar X=5.97$ represent the sample mean for the length

$s=0.4$ represent the sample standard deviation

$n=93$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value for the test

System of hypothesis

We need to conduct a hypothesis in order to check if the lathe is in perfect adjustment (6cm), then the system of hypothesis would be:

Null hypothesis: $\mu = 6$

Alternative hypothesis: $\mu \neq 6$

since we don't know the population deviation the statistic is:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing in formula (1) we got:

$t=\frac{5.97-6}{\frac{0.4}{\sqrt{93}}}=-0.723$

E. -0.723

P value

The degrees of freedom are given by:

$df=n-1=93-1=92$

Since is a two tailed test the p value would be:

$p_v =2*P(t_{(92)}$

Since the p value is very high we don't have enough evidence to conclude that the true mean for the lengths is different from 6 cm.

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Read 2 more answers

Find the value of the polynomial 5x – 4x^2 + 3 at x = 2 and x = –1​

Find the value of the polynomial 5x – 4x^2 + 3 at x = 2 and x = –1