Carole earns $2.50 for each gadget she completes. What was her gross pay if she completed 145 gadgetslast week?

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Answer 1

Answer: $2.50$ per gadget multiplied by $145$ gadgets is a total of $362.50$ .

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True or false: In (r, theta), the value of r can be negative.

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The answer to the mathematics question presented above is 'True'. It is correct to say that in (r theta), the value of r can be negative. A negative radius can be used when it comes to graphing a "polar'' function. Thus, the answer to the question is 'true'.

In the picture below

If eric practiced the piano 8hrs a week, if he practices 1/4 of that how many hours is that?

Answers

divide 8 by 4 and you get 2 hours

Work:

It is fourths, so divide
8 ÷ 4
That gives you 2

Answer:

2 hours

Which function will have a y-intercept at -1 and an amplitude of 2?

Answers

The first and 3rd choices are just sin and cos waves, with amplitude of 1 .
So it has to be the 2nd or 4th choice.

The y-intercept is the point on the graph where x=0.
So if the y-intercept is -1, that just means that f(0)=-1 .
So let's test the two choices:

   f(x) = -2 sin(x) - 1

   f(0) = -2 sin(0) - 1
sin(0) = 0 so
   f(0) = -2 (0) - 1 .
That looks like it works.
Just to be sure, let's check the other possible choice.

f(x) = -2 cos(x) - 1

   f(0) = -2 cos(0) - 1

cos(0) = 1 so f(0) = -2 (1)    - 1 = -3 .

Beautiful ! The y-intercept of choice-2 is -1,
and the y-intercept of choice-4 is -3. The answer is choice-2.

Answer:

the answer is b

Step-by-step explanation:

Im just typing this so it knows that I properly explained

3.30 Measurements of scientific systems are always subject to variation, some more than others. There are many structures for measurement error, and statisticians spend a great deal of time modeling these errors. Suppose the measurement error X of a certain physical quantity is decided by the density function f(x) = k(3 − x2), −1 ≤ x ≤ 1, 0, elsewhere. (a) Determine k that renders f(x) a valid density function. (b) Find the probability that a random error in measurement is less than 1/2. (c) For this particular measurement, it is undesirable if the magnitude of the error (i.e., |x|) exceeds 0.8. What is the probability that this occurs?

Answers

Answer:

a) k should be equal to 3/16 in order for f to be a density function.

b) The probability that the measurement of a random error is less than 1/2 is 0.7734

c) The probability that the magnitude of a random error is more than 0.8 is 0.164

Step-by-step explanation:

a) In order to find k we need to integrate f between -1 and 1 and equalize the result to 1, so that f is a density function.

$1 = k \int\limits^1_(-1) {(3-x^2)} \, dx = k * (3x-(x^3)/(3))|_(x=-1)^(x = 1) = k*[(3-1/3) - (-3 + 1/3)] = 16k/3$

16k/3 = 1

k = 3/16

b) For this probability we have to integrate f between -1 and 0.5 (since f takes the value 0 for lower values than -1)

$P(X < 1/2) = \int\limits^(0.5)_(-1) {(3)/(16)(3-x^2)} \, dx = (3)/(16) [(3x-(x^3)/(3)) |_(x=-1)^(x=0.5)] =(3)/(16) *(1.458333 - (-3+1/3)) = 0.7734$

c) For |x| to be greater than 0.8, either x>0.8 or x < -0.8. We should integrate f between 0.8 and 1, because we want values greater than 0.8, and f is 0 after 1; and between -1 and 0.8.

$P(|X| > 0.8) = \int\limits^(-0.8)_(-1) {(3)/(16)*(3-x^2)} \, dx + \int\limits^(1)_(0.8) {(3)/(16)*(3-x^2)} \, dx =\n (3)/(16) (3x-(x^3)/(3))|_(x=-1)^(x=-0.8) + (3)/(16) (3x-(x^3)/(3))|_(x=0.8)^(x=1) = 0.082 + 0.082 = 0.164$

(a) The value of k that makes f(x) a valid density function is k = 1/6.

(b) The probability that a random error in measurement is less than 1/2 is 3/4.

(a) To make the given function f(x) a valid probability density function, it must satisfy the following conditions:

The function must be non-negative for all x: f(x) ≥ 0.

The total area under the probability density function must equal 1: ∫f(x)dx from -1 to 1 = 1.

Given $f(x) = k(3 - x^2)$ , -1 ≤ x ≤ 1, and f(x) = 0 elsewhere, let's find the value of k that satisfies these conditions.

Non-negativity: The function is non-negative for -1 ≤ x ≤ 1, so we have $k(3 - x^2)$ ≥ 0 for -1 ≤ x ≤ 1. This means that k can be any positive constant.

Total area under the probability density function: To find the value of k, integrate f(x) over the interval [-1, 1] and set it equal to 1:

∫[from -1 to 1] $k(3 - x^2)dx$ = 1

∫[-1, 1] $(3k - kx^2)dx$ = 1

Now, integrate the function:

$[3kx - (kx^3/3)]$ from -1 to 1 = 1

$[(3k(1) - (k(1^3)/3)) - (3k(-1) - (k(-1^3)/3))] = 1$

Simplify:

[3k - k/3 + 3k + k/3] = 1

6k = 1

k = 1/6

So, the value of k that makes f(x) a valid density function is k = 1/6.

(b) To find the probability that a random error in measurement is less than 1/2, you need to calculate the integral of f(x) from -1/2 to 1/2:

P(-1/2 ≤ X ≤ 1/2) = ∫[from -1/2 to 1/2] f(x)dx

P(-1/2 ≤ X ≤ 1/2) = ∫[-1/2, 1/2] (1/6) $(3 - x^2)dx$

Now, integrate the function:

$(1/6) [3x - (x^3/3)]$ from -1/2 to 1/2

$[(1/6)(3(1/2) - ((1/2)^3/3)) - (1/6)(3(-1/2) - ((-1/2)^3/3))]$

Simplify:

(1/6)[(3/2 - 1/24) - (-3/2 + 1/24)]

(1/6)[(9/8) + (9/8)]

(1/6)(18/8)

(3/4)

So, the probability that a randomerror in measurement is less than 1/2 is 3/4.

(c) To find the probability that the magnitude of theerror (|x|) exceeds 0.8, you need to calculate the probability that |X| > 0.8. This is the complement of the probability that |X| ≤ 0.8, which you can calculate as:

P(|X| > 0.8) = 1 - P(|X| ≤ 0.8)

P(|X| > 0.8) = 1 - P(-0.8 ≤ X ≤ 0.8)

We already found P(-0.8 ≤ X ≤ 0.8) in part (b) to be 3/4, so:

P(|X| > 0.8) = 1 - 3/4

P(|X| > 0.8) = 1/4

So, the probability that the magnitude of the error exceeds 0.8 is 1/4.

To Learn more about probability here:

brainly.com/question/13604758

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Using the square root property: How would I solve this equation- (2x-3) Squared =18

Answers

$(2x-3)^2=18\ \ \ \ | make\ \sqrt{}\n\n|2x-3|=√(18)\n\n2x-3=√(18)\ \ \ or\ \ \ -2x+3=√(18)\n\n2x=√(18)+3\ \ \ \ or\ \ \ \ -2x=√(18)-3\n\nx=(√(9*2)+3)/(2)\ \ \ or\ \ \ \ x=(√(9*2)-3)/(-2)\n\nx=(3√(2)+3)/(2)\ \ \ or\ \ \ \ x=(-3√(2)+3)/(2)$

$(2x-3)^2=18\n2x-3=√(18) \vee 2x-3=-√(18)\n2x=3+3√(2) \vee 2x=3-3√(2)\nx=(3+3√(2))/(2) \vee x=(3-3√(2))/(2)\n$

If x and y are negative integers and x – y = 1,
what is the least possible value for xy?

Answers

x=y+1
Least possible value for y(y+1)? Well, it appears for y halfway between the roots (you know what a parabola is, don't you?). That y is -0.5, so let's calculate -0.5*0.5, which is -0.25.
The least possible value is $-\frac{1}4$
Well, not really :)) The two values must be both negative, so we can't take one of them to be 0.5! As we lower the values, however, the product grows, reaching 0 when x=0 and y=-1, and further increasing as x and y lower. Thus, we would actually take x=0 and y=-1.

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