4x-5.2y + 6y + 7.9x

A sample of n = 5 scores has a mean of m = 8. one new score is added to the sample and the new mean is calculated to be m = 9. w

Yuri [45]

If the 5 scores have a mean of 8, then their total sum would be 8*5 = 40

Now if one score if added (let's call this score x), there are 6 scores and the mean changes to 9, thus:
(40 + x)/6 = 9
40 + x = 54
x = 14

6 0

3 years ago

Read 2 more answers

One week Isabella bought 2 bags of apples 3 bags of oranges. Next week she bought 1 bag of apples and 2 bags of oranges what exp

TEA [102]

Answer:3x+5y

Step-by-step explanation:

5 0

3 years ago

Please help!!As soon as possible!!

pychu [463]

Answer: Their equations have different y-intercept but the same slope

Step-by-step explanation:

Since, the slope of the line passes through the points (2002,p) and (2011,q) is,

$m_1 = \frac{q-p}{2011-2002} = \frac{q-p}{9}$

Similarly, the slope of the line asses through the points (2,p) and (11,q) is,

$m_2 = \frac{q-p}{11-2} = \frac{q-p}{9}$

Since, $m_1 = m_2$

Hence, both line have the same slope.

Now, the equation of the line one having slope $m_1$ and passes through the point (2002,p) is,

$y - p = \frac{q-p}{9}(x-2002)$

Put x = 0 in the above equation,

We get, $y = \frac{2000(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2000(p-q)}{9}+p)$

Also, the equation of second line having slope $m_2$ and passes through the point (2,p)

$y-p=\frac{q-p}{9}(x-2)$

Put x = 0 in the above equation,

We get, $y = \frac{2(p-q)}{9}+p$

The y-intercept of the line one is $(0, \frac{2(p-q)}{9}+p)$

Thus, both line have the different y-intercepts.

⇒ Third option is correct.

6 0

3 years ago

A chocolate factory made 708.6 pounds of dark chocolate in 6 days. How much dark chocolate, on average, did the factory make per

OLEGan [10]

Answer:

118.1

Step-by-step explanation:

You need to divide 708.6 by 6 to get the amount of chocolate in one day

3 0

3 years ago

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x4 ln(x) (a) Find the interval on which f is incre

Ainat [17]

Answer: (a) Interval where f is increasing: (0.78,+∞);

Interval where f is decreasing: (0,0.78);

(b) Local minimum: (0.78, - 0.09)

(c) Inflection point: (0.56,-0.06)

Interval concave up: (0.56,+∞)

Interval concave down: (0,0.56)

Step-by-step explanation:

(a) To determine the interval where function f is increasing or decreasing, first derive the function:

f'(x) = $\frac{d}{dx}$ [ $x^{4}ln(x)$ ]

Using the product rule of derivative, which is: [u(x).v(x)]' = u'(x)v(x) + u(x).v'(x),

you have:

f'(x) = $4x^{3}ln(x) + x_{4}.\frac{1}{x}$

f'(x) = $4x^{3}ln(x) + x^{3}$

f'(x) = $x^{3}[4ln(x) + 1]$

Now, find the critical points: f'(x) = 0

$x^{3}[4ln(x) + 1]$ = 0

$x^{3} = 0$

x = 0

and

$4ln(x) + 1 = 0$

$ln(x) = \frac{-1}{4}$

x = $e^{\frac{-1}{4} }$

x = 0.78

To determine the interval where f(x) is positive (increasing) or negative (decreasing), evaluate the function at each interval:

interval x-value f'(x) result

0<x<0.78 0.5 f'(0.5) = -0.22 decreasing

x>0.78 1 f'(1) = 1 increasing

With the table, it can be concluded that in the interval (0,0.78) the function is decreasing while in the interval (0.78, +∞), f is increasing.

Note: As it is a natural logarithm function, there are no negative x-values.

(b) A extremum point (maximum or minimum) is found where f is defined and f' changes signs. In this case:

Between 0 and 0.78, the function decreases and at point and it is defined at point 0.78;
After 0.78, it increase (has a change of sign) and f is also defined;

Then, x=0.78 is a point of minimum and its y-value is:

f(x) = $x^{4}ln(x)$

f(0.78) = $0.78^{4}ln(0.78)$