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3 years ago

Please help solve the inequality a+4/2 > 16

Mathematics

2 answers:

wel3 years ago

5 0

Answer:123

Step-by-step explanation:

serg [7]3 years ago

3 0

Answer:

a>28

Step-by-step explanation:

You might be interested in

Suppose that a sample of size 100 is to be drawn from a population with standard deviation L0. a. What is the probability that t

NARA [144]

Answer:

The probability that the sample mean will lie within 2 values of μ is 0.9544.

Step-by-step explanation:

Here

the sample size is given as 100
the standard deviation is 10

The probability that the sample mean lies with 2 of the value of μ is given as

$P(| \bar{X}-\mu|$

Here converting the values in z form gives

$P(-2$

Substituting values

$P(-2$

From z table

$P(z\leq 2)=0.9772\\P(z\leq -2)=0.0228\\P(-2\leq z\leq 2)=P(z\leq 2)-P(z\leq -2)\\P(-2\leq z\leq 2)=0.9772-0.0228\\P(-2\leq z\leq 2)=0.9544\\$

So the probability that the sample mean will lie within 2 values of μ is 0.9544.

5 0

3 years ago

PLEASE HELP, GOOD ANSWERS GET BRAINLIEST. +40 POINTS WRONG ANSWERS GET REPORTED

MA_775_DIABLO [31]

1. Ans:(A) 123

Given function: $f(x) = 8x^2 + 11x$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2 + 11x)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(8x^2) + \frac{d}{dx}(11x)$
=> $\frac{d}{dx} f(x) = 2*8(x^{2-1}) + 11$
=> $\frac{d}{dx} f(x) = 16x + 11$

Now at x = 7:
$\frac{d}{dx} f(7) = 16(7) + 11$

=> $\frac{d}{dx} f(7) = 123$

2. Ans:(B) 3

Given function: $f(x) =3x + 8$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(3x + 8)$
=> $\frac{d}{dx} f(x) = \frac{d}{dx}(3x) + \frac{d}{dx}(8)$
=> $\frac{d}{dx} f(x) = 3*1 + 0$
=> $\frac{d}{dx} f(x) = 3$

Now at x = 4:
$\frac{d}{dx} f(4) = 3$ (as constant)

=>Ans: $\frac{d}{dx} f(4) =$ 3

3. Ans:(D) -5

Given function: $f(x) = \frac{5}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{5}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(5x^{-1})$
=> $\frac{d}{dx} f(x) = 5*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = -5x^{-2}$

Now at x = -1:
$\frac{d}{dx} f(-1) = -5(-1)^{-2}$

=> $\frac{d}{dx} f(-1) = -5 *\frac{1}{(-1)^{2}}$
=> Ans: $\frac{d}{dx} f(-1) = -5$

4. Ans:(C) 7 divided by 9

Given function: $f(x) = \frac{-7}{x}$
The derivative would be:
$\frac{d}{dx} f(x) = \frac{d}{dx}(\frac{-7}{x})$
or
$\frac{d}{dx} f(x) = \frac{d}{dx}(-7x^{-1})$
=> $\frac{d}{dx} f(x) = -7*(-1)*(x^{-1-1})$
=> $\frac{d}{dx} f(x) = 7x^{-2}$

Now at x = -3:
$\frac{d}{dx} f(-3) = 7(-3)^{-2}$

=> $\frac{d}{dx} f(-3) = 7 *\frac{1}{(-3)^{2}}$
=> Ans: $\frac{d}{dx} f(-3) = \frac{7}{9}$

5. Ans:(C) -8

Given function:
$f(x) = x^2 - 8$

Now if we apply limit:
$\lim_{x \to 0} f(x) = \lim_{x \to 0} (x^2 - 8)$

=> $\lim_{x \to 0} f(x) = (0)^2 - 8$
=> Ans: $\lim_{x \to 0} f(x) = - 8$

6. Ans:(C) 9

Given function:
$f(x) = x^2 + 3x - 1$

Now if we apply limit:
$\lim_{x \to 2} f(x) = \lim_{x \to 2} (x^2 + 3x - 1)$

=> $\lim_{x \to 2} f(x) = (2)^2 + 3(2) - 1$
=> Ans: $\lim_{x \to 2} f(x) = 4 + 6 - 1 = 9$

7. Ans:(D) doesn't exist.

Given function: $f(x) = -6 + \frac{x}{x^4}$
In this case, even if we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

Check:
$f(x) = -6 + \frac{x}{x^4} \\ f(x) = -6 + \frac{1}{x^3} \\ f(x) = \frac{-6x^3 + 1}{x^3} \\ Rationalize: \\ f(x) = \frac{-6x^3 + 1}{x^3} * \frac{x^{-3}}{x^{-3}} \\ f(x) = \frac{-6x^{3-3} + x^{-3}}{x^0} \\ f(x) = -6 + \frac{1}{x^3} \\ Same$

If you apply the limit, answer would be infinity.

8. Ans:(A) Doesn't Exist.

Given function: $f(x) = 9 + \frac{x}{x^3}$
Same as Question 7
If we try to simplify it algebraically, there would ALWAYS be x power something (positive) in the denominator. And when we apply the limit approaches to 0, it would always be either + infinity or -infinity. Hence, Limit doesn't exist.

9, 10.
Please attach the graphs. I shall amend the answer. :)

11. Ans:(A) Doesn't exist.

First We need to find out: $\lim_{x \to 9} f(x)$ where,
$f(x) = \left \{ {{x+9, ~~~~~x \textless 9} \atop {9- x,~~~~~x \geq 9}} \right.$

If both sides are equal on applying limit then limit does exist.

Let check:
If x $\textless$ 9: answer would be 9+9 = 18
If x $\geq$ 9: answer would be 9-9 = 0

Since both are not equal, as $18 \neq 0$ , hence limit doesn't exist.

12. Ans:(B) Limit doesn't exist.

Find out: $\lim_{x \to 1} f(x)$ where,

$f(x) = \left \{ {{1-x, ~~~~~x \textless 1} \atop {x+7,~~~~~x \textgreater 1} } \right. \\ and \\ f(x) = 8, ~~~~~ x=1$

If all of above three are equal upon applying limit, then limit exists.

When x < 1 -> 1-1 = 0
When x = 1 -> 8
When x > 1 -> 7 + 1 = 8

ALL of the THREE must be equal. As they are not equal. 0 $\neq$ 8; hence, limit doesn't exist.

13. Ans:(D) -∞; x = 9

f(x) = 1/(x-9).

Table:

x f(x)=1/(x-9)

----------------------------------------

8.9 -10

8.99 -100

8.999 -1000

8.9999 -10000

9.0 -∞

Below the graph is attached! As you can see in the graph that at x=9, the curve approaches but NEVER exactly touches the x=9 line. Also the curve is in downward direction when you approach from the left. Hence, -∞, x =9 (correct)

14. Ans: -6

s(t) = -2 - 6t

Inst. velocity = $\frac{ds(t)}{dt}$

Therefore,

$\frac{ds(t)}{dt} = \frac{ds(t)}{dt}(-2-6t) \\ \frac{ds(t)}{dt} = 0 - 6 = -6$

At t=2,

Inst. velocity = -6

15. Ans: +∞, x =7

f(x) = 1/(x-7)^2.

Table:

x f(x)= 1/(x-7)^2

--------------------------

6.9 +100

6.99 +10000

6.999 +1000000

6.9999 +100000000

7.0 +∞

Below the graph is attached! As you can see in the graph that at x=7, the curve approaches but NEVER exactly touches the x=7 line. The curve is in upward direction if approached from left or right. Hence, +∞, x =7 (correct)

-i

7 0

3 years ago

Read 2 more answers

Someone help please i am so bad at point slope stuff

Fynjy0 [20]

<span>1) Write the point-slope form of the equation of the horizontal line that passes through the point (2, 1). y = 1/2x

2)Write the point-slope form of the equation of the line that passes through the points (6, -9) and (7, 1).
m = (-9 - 1) / (6 - 7) = -10/-1 = 10
y + 9 = 10 (x - 6)
y = 10x - 69

3) A line passes through the point (-6, 6) and (-6, 2). In two or more complete sentences, explain why it is not possible to write the equation of the given line in the traditional version of the point-slope form of a line.

4)Write the point-slope form of the equation of the line that passes through the points (-3, 5) and (-1, 4).
m = (5 - 4) / (-3 - -1) = 1/-2
y - 5 = (-1/2) (x +3)
y = (-1/2)x + 7/2

5) Write the point-slope form of the equation of the line that passes through the points (6, 6) and (-6, 1).
m = (6-1)/(6 - -6) = 5 / 12
y - 6 = (5/12) (x-6)
y = (5/12)x + 17 / 2

6) Write the point-slope form of the equation of the line that passes through the points (-8, 2) and (1, -4).
m = (2 - -4) / (-8 -1) = 6 / -7
y - 2 = (-6/7) (x + 8)
y = (-6/7)x - 50 / 7

7) Write the point-slope form of the equation of the line that passes through the points (5, -9) and (-6, 1).
m = (-9 - 1) / (5 - -6) = -10 / 11
y + 9 = (-10 / 11) (x - 5)
y = (-10 / 11)x -49/11

</span>

3 0

4 years ago

A fruit stand sells apples, peaches and tomatoes. Today, they sold 4 apples for every 5 peaches. They sold 2 peaches for every 3

Brrunno [24]

Answer:

He sold 32 apples, 40 peaches and 60 apples.

Step-by-step explanation:

We are given that

4 apples are sold every 5 peaches and

3 tomatoes are sold every 2 peaches.

i.e. the ratio is apples: peaches = 4 : 5 ...... (1)

and the ratio peaches: tomatoes = 2 : 3 ...... (2)