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s344n2d4d5 [400]
1 year ago
9

Alan has to pay Robin $48, but he has only $36 with him. Alan overdraws the remaining amount from the bank to pay Robin. After p

aying Robin, what will Alan's account balance be?
Mathematics
2 answers:
lianna [129]1 year ago
7 0

Answer:

The answer would be $-12.

Step-by-step explanation:

Alan has to overdraw his account by $12. So his account balance will be -$12.

Rashid [163]1 year ago
6 0
I assume his account balance will become negative. With no overdraft fees in the question, then his account balance should be $-12. 
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Logan wants to move to a new city. He gathered graphs of temperatures for two different cities. Which statements about the data
Bad White [126]

Answer:

The correct options are;

Therefore, City A is likely to have temperatures that remain fairly constant all year round because it has a compact interquartile range compared to that of City B

City B is likely to have more extreme temperatures with colder days in the winter and hotter days in the summer because the range is greater than that of A

Step-by-step explanation:

Here we have for City A

Maximum - Minimum = 10

Interquartile range =3

City B

Maximum - Minimum = 18.5

Interquartile range =9.5

Therefore, City A is likely to have temperatures that remain fairly constant all year round because it has a compact interquartile range compared to that of City B

City B is likely to have more extreme temperatures with colder days in the winter and hotter days in the summer because the range is greater than that of A.

7 0
2 years ago
Read 2 more answers
Samara is adjusting a satellite because she finds it is not focusing the income radio waves perfectly. The shape of her satellit
gogolik [260]

Answer:

\boxed{\text{(6, 3)}}

Step-by-step explanation:

The conic form of the equation for a sideways parabola is

(y - k)² = 4p(x - h)

The focus is at (h + p, k)

The equation of Samara's parabola is

(y - 3)² = 8(x - 4)

h = 4

p = 8/4 = 2

k = 3

h + p = 6

So, the focus point of the satellite dish is at

\boxed{\textbf{(6, 3)}}

8 0
1 year ago
A given set of values is found to be a normal distribution with a mean of 140 and a standard deviation of 18.0. Find the value t
Alisiya [41]

Answer:

The value that is greater than 45% of the data values is approximately 137.84.

Step-by-step explanation:

The key is transforming values from this distribution to a z-score range and finding the corresponding value using a z-score table.

We are looking for a value x which attains a critical z-score that corresponds to the (100-45)%=55-th percentile:

z_{0.55} = \frac{x-\mu}{\sigma}=\frac{x-140}{18}\implies x = 18\cdot z_{0.55}+140

The critical z value (from z-score table, online) is: -0.12, so:

x = 18\cdot z_{0.55}+140=18\cdot(-0.12)+140\approx137.84

The value that is greater than 45% of the data values is approximately 137.84.


5 0
2 years ago
Read 2 more answers
Evaluate <br> 17.3<br> % of <br> 45.94<br> km<br> Give your answer rounded to 2 DP.
Katyanochek1 [597]

Answer: 17.3% of 45.94 km is 7.95 km

4 0
2 years ago
The graphs of the quadratic functions f(x) = 6 – 10x2 and g(x) = 8 – (x – 2)2 are provided below. Observe there are TWO lines si
natta225 [31]

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The line simultaneously tangent to both graphs having the LARGEST slope has equation

(b) The other line simultaneously tangent to both graphs has equation,

Solution:

- Find the derivatives of the two functions given:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Since, the derivative of both function depends on the x coordinate. We will choose a point x_o which is common for both the functions f(x) and g(x). Point: ( x_o , g(x_o)) Hence,

                                g'(x_o) = -2*(x_o -2)

- Now compute the gradient of a line tangent to both graphs at point (x_o , g(x_o) ) on g(x) graph and point ( x , f(x) ) on function f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now the gradient of the line computed from a point on each graph m must be equal to the derivatives computed earlier for each function:

                                m = f'(x) = g'(x_o)

- We will develop the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Form factors:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o   ,     x_o = 10x + 2    

- For x_o = 10x + 2  ,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Solve the quadratic equation above:

                                 x = -0.0574, -0.387      

- Largest slope is at x = -0.387 where equation of line is:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- Other tangent line:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036

6 0
1 year ago
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