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3 years ago

Add the polynomias:(8x^2-5x+7)+(7x^3-3x)

Mathematics

2 answers:

mezya [45]3 years ago

3 0

Answer:

$7x^{3} + 8x^{2} -8x + 7$

Step-by-step explanation:

$(8x^{2} - 5x + 7) + (7x^{3} - 3x) = 8x^{2} - 5x + 7 + 7x^{3} - 3x = 7x^{3} + 8x^{2} - 5x - 3x + 7 = 7x^{3} + 8x^{2} -8x + 7$

xxMikexx [17]3 years ago

3 0

Answer:

7x³ + 8x² -8x + 7

Step-by-step explanation:

8x² - 5x +7 + 7x³ - 3x

7x³ + 8x² + (-5x -3x ) + 7

7x³ + 8x² -8x + 7

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3 years ago

Piece made in 32min factory cut the time by 8% then cut time by 5% how long does it take to make the piece now in min

DiKsa [7]

Answer:

27.968 mins

Step-by-step explanation:

1) we need to find 8% of 32:

32*0.08=2.56

so 8% of 32 is 2.56 mins.

we need to subtract 2.56 by 32 since its says the times was cut by 8%

32-2.56=29.44

now 5% of 29.44:

29.44*0.05=1.472

1.472 mins

29.44-1.472=27.968

Hope this helps!

4 0

3 years ago

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha }{2} } = 1.96$

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

Then the critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

$\= x - E < \mu < \= x + E$

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

3 years ago

Twenty-four students tried out for the cheerleading squad. Only 5/6 of the

IceJOKER [234]

Step-by-step explanation:

Total no. of students tried out for the cheerleading squad = 24

Only 5/6 of the students will be chosen, therefore, no.s of students will be chosen for the squad =

$\frac{5}{6} \times 24$