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3 years ago

A number cube is rolled. What is the probability of rolling a 2,3, or 5?

1 answer:

4 0

Well if the number cube goes up to 6 like most number cubes, then the probability would be 3/6 because there are 3 numbers that you want to roll (2, 3, and 5) out of 6 total numbers (1-6). Then 3/6 can be further simplified to 1/2. So the answer should be 1/2.

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Please help me with this. This problem is actually making my life worst at the moment. Please, someone, help me.

Xelga [282]

OMG !!!!!!! I HAVE THE SAME QUESTION TO ASK

7 0

3 years ago

Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas

o-na [289]

Answers:

(a) f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing at $(-2,2)$ .

(c) f is concave up at $(2, \infty)$

(d) f is concave down at $(-\infty, 2)$

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

$f'(x) = 15x^2 - 60
$

So,

$
f'(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \ 0} \text{ (1)}$

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

$f'(x) \ \textgreater \ 0 \Leftrightarrow (x - 2)(x + 2) \ \textgreater \ 0$

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at $(-\infty,-2) \cup (2,\infty)$ .

(b) f is decreasing only when the derivative of f is negative. Since

$f'(x) = 15x^2 - 60$

Using the similar computation in (a),

$f'(x) \ \textless \ \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \ 0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \ 0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \ 0} \text{ (2)}$

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

$f''(x) = 30x - 60$

Since,

$f''(x) \ \textgreater \ 0
\\
\\ \Leftrightarrow 30x - 60 \ \textgreater \ 0
\\
\\ \Leftrightarrow 30(x - 2) \ \textgreater \ 0
\\
\\ \Leftrightarrow x - 2 \ \textgreater \ 0
\\
\\ \Leftrightarrow \boxed{x \ \textgreater \ 2}$

Therefore, f is concave up at $(2, \infty)$ .

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

$f''(x) = 30x - 60$

Using the similar computation in (c),

$f''(x) \ \textless \ 0 
\\ \\ \Leftrightarrow 30x - 60 \ \textless \ 0 
\\ \\ \Leftrightarrow 30(x - 2) \ \textless \ 0 
\\ \\ \Leftrightarrow x - 2 \ \textless \ 0 
\\ \\ \Leftrightarrow \boxed{x \ \textless \ 2}$

Therefore, f is concave down at $(-\infty, 2)$ .

3 0

3 years ago

in a study evaluating the relationship between stress and muscle cramps, half the subjects are randomly assigned to be exposed t

stiks02 [169]

The study creates stress in a(0ne) way that might affect muscle cramps.

What is muscle cramps ?

Overuse of a muscle, dehumidification, muscle strain or simply holding a position for a prolonged period can beget a muscle cramp. In numerous cases, still, the cause is not known.
Although utmost muscle cramps are inoffensive, some may be related to an beginning medical condition, similar as shy blood force.
Cramps can have causes that are not due to underpinning complaint. exemplifications include dehumidification, emphatic exercise or lack of muscle use. Muscle cramps are unforeseen muscle condensation.
Also called muscle spasms or charley nags, a muscle cramp can be a common symptom of numerous effects, like exercise strain or a medical condition. Muscle cramps generally go down without treatment and can be watched for at home.

The study creates stress in a(0ne) way that might affect muscle cramps. If an elevator falls, fleetly and stops suddenly, that could affect muscle cramps more because of the physical action than by the stress it convinced. I don’t suppose this trial would help answer whether there's a unproductive relationship between increases stress and muscle cramps.

Learn more about muscle cramp here:

brainly.com/question/19064082

#SPJ4

7 0

1 year ago

Marukh can wash the car in 10 minutes. steven can wash the car in 60 minutes. if marukh and steven work together, how long will

yKpoI14uk [10]

I believe that the answer is 35 minutes
as that is their average in minutes for how long it takes to wash the car.

60 + 10 = 70
70 / 2 = 35

5 0

2 years ago

Head movement evaluations are important because disabled individuals may be able to operate communications aids using head motio

earnstyle [38]

Answer:

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.307 -0}{\frac{7.798}{\sqrt{14}}}=0.6271$

df=n-1=14-1=13

$p_v =P(t_{(13)}>0.6271) =0.2707$

Step-by-step explanation:

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value CL , y = test value CO

x: 57.8 35.7 54.5 56.8 51.1 70.8 77.3 51.6 54.7 63.6 59.2 59.2 55.8 38.5

y: 44.7 52.1 60.2 52.7 47.2 65.6 71.4 48.8 53.1 66.3 59.8 47.5 64.5 34.4

The system of hypothesis for this case are:

Null hypothesis: $\mu_x- \mu_y \leq 0$

Alternative hypothesis: $\mu_x -\mu_y >0$

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: 13.1, -16.4,-5.7,4.1,3.9,5.2,5.9,2.8,1.6,-2.7,-0.6,11.7,-8.7,4.1

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{29}{8}=1.307$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =7.798$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{1.307 -0}{\frac{7.798}{\sqrt{14}}}=0.6271$

The next step is calculate the degrees of freedom given by:

$df=n-1=14-1=13$

Now we can calculate the p value, since we have a right tailed test the p value is given by:

$p_v =P(t_{(13)}>0.6271) =0.2707$

So the p value is higher than the significance level provided of 0.01, so then we can conclude that we FAIL to reject the null hypothesis that the difference mean between after and before is higher or equal than 0. So we can't conclude that the mean neck rotation is greater in the clockwise direction than in the counterclockwise direction at 1% of significance.

8 0

3 years ago