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3 years ago

7

If an equation has two roots -4 and 7, what could be the quadratic equation?

1 answer:

aleksklad [387]3 years ago

8 0

Answer:

f(x) = x^2 - 3x - 28

Step-by-step explanation:

The root -4 corresponds to the factor (x + 4); the root 7 corresponds to (x - 7). The equation is thus f(x) = (x + 4)(x - 7) = x^2 - 3x - 28, or

f(x) = x^2 - 3x - 28

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What’s 3x squared plus 9x-2

KengaRu [80]

Answer:

The expression is not factorable with rational numbers.

3x2+9x−2

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3 years ago

The histogram to the right represents the weights (in pounds) of members of a certain high-school programming team. What is th

AVprozaik [17]

Answer:

Class width=20

Lower class limit of first class=100

Upper class limit of first class=120

Step-by-step explanation:

The class width can be calculated by taking the difference of two consecutive upper class limits or lower class limits.

Now we take any two consecutive upper class limits or lower class limits from classes 100-120,120-140,140-160,160-180,180-200,200-220,220-240.

We take upper class limits of first and second class i.e. 100 and 120.

Class width=120-100=20

Class width=20

The approximate lower and upper class limits of the first class from classes 100-120,120-140,140-160,160-180,180-200,200-220,220-240 are 100 and 120.

Class limits for first class is 100-120.

Lower class limit of first class=100

Upper class limit of first class=120

4 0

3 years ago

42how many times gets you 204

IrinaVladis [17]

Answer:

8568

Step-by-step explanation:

multiply 42 and 204 and you get 8568

7 0

3 years ago

Read 2 more answers

A meat inspector has randomly selected 30 packs of 95% lean beef. The sample resulted in a mean of 96.2% with a sample standard

kirza4 [7]

Answer:

a

The upper bound of the 99% prediction level is $98.2$

b

The 95% confidence interval is $9.7383 < \mu < 10.2617$

Step-by-step explanation:

Considering first question

From the question we are told that

The sample size is n = 30

The sample mean is $\= x = 96.2\%$

The standard deviation is $s = 0.8\%$

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 30 - 1$

=> $df = 29$

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\alpha , 29} = 2.462$

Generally the 99% prediction level is mathematically represented as

$\= x \pm [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

Generally the upper bound of the 99% prediction level is mathematically represented as

$\= x + [(t_{\alpha , df }) * s * (\sqrt{1 + \frac{1}{ n} } )}]$

=> $96.2 + (2.462 ) * 0.8 * (\sqrt{1 + \frac{1}{ 30} } )}]$

=> $98.2$

Considering second question

Generally the sample is mathematically represented as

$\= x = \frac{\sum x_i}{n}$

=> $\= x = \frac{ 9.8 + 10.2 + \cdots +9.6 }{7}$

=> $\= x = 10$

Generally the standard deviation is mathematically represented as

$\sigma = \sqrt{ \frac{ \sum ( x_ i - \= x)}{n-1} }$

=> $\sigma = \sqrt{ \frac{ ( 9.8 -10)^2 + ( 10.2 -10)^2 + \cdots + ( 9.6 -10)^2 }{7-1} }$

=> $\sigma = 0.283$

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 7- 1$

=> $df = 6$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the t distribution table the critical value of at a degree of freedom of is

$t_{\frac{\alpha }{2} , 6 } = 2.447$

Generally the margin of error is mathematically represented as

$E = t_{\frac{\alpha }{2} , 6 } * \frac{\sigma }{\sqrt{n} }$

=> $E =2.447* \frac{0.283 }{\sqrt{7} }$

=> $E =0.2617$

Generally 95% confidence interval is mathematically represented as

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

=> $10 -0.2617 < \mu < 10 + 0.2617$

=> $9.7383 < \mu < 10.2617$

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3 years ago

Find the value of x log2x-log(x-3)=1

rusak2 [61]

Answer:

15/4

Step-by-step explanation:

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3 years ago

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