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3 years ago

Regina charges c dollars per hour to babysit. If she increases her rate by 15%, which

Mathematics

2 answers:

Inga [223]3 years ago

8 0

Answer:

C. c+0.15c

Step-by-step explanation:

Raising her rate by 15% is equivalent to adding 15% of what she charged before to the original price.

Pavlova-9 [17]3 years ago

6 0

Yes. C is the correct answer.

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Four times the sum of 14 and c is a squared

AlekseyPX

$4\times(14+c^2)=4\times14+4\times c^2=56+4c^2$

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3 years ago

An important problem in industry is shipment damage. A windshield factory ships its product by truck and determines that it cann

ioda

Answer:

Option f) None of these

Step-by-step explanation:

We are given the following data set:

Sample size, n = 12

Sample mean = 9.4

Sample standard deviation = 0.64

Confidence interval:

$\bar{x} \pm t_{critical}\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 11 and at}~\alpha_{0.01} = \pm 3.1058$

$9.4 \pm 3.1058(\frac{0.64}{\sqrt{12}} ) = 9.4 \pm 0.574 = (8.826,9.974)$

5 0

3 years ago

(3+1)²·3 plz help me

xxMikexx [17]

Answer:

Step-by-step explanation:

( 3+1)² * 3

= 4² * 3

= 16 * 3

= 48

6 0

3 years ago

Help me with trigonometry

poizon [28]

Answer:

See below

Step-by-step explanation:

It has something to do with the Weierstrass substitution, where we have

$\int\, f(\sin(x), \cos(x))dx = \int\, \dfrac{2}{1+t^2}f\left(\dfrac{2t}{1+t^2}, \dfrac{1-t^2}{1+t^2} \right)dt$

First, consider the double angle formula for tangent:

$\tan(2x)= \dfrac{2\tan(x)}{1-\tan^2(x)}$

Therefore,

$\tan\left(2 \cdot\dfrac{x}{2}\right)= \dfrac{2\tan(x/2)}{1-\tan^2(x/2)} = \tan(x)=\dfrac{2t}{1-t^2}$

Once the double angle identity for sine is

$\sin(2x)= \dfrac{2\tan(x)}{1+\tan^2(x)}$

we know $\sin(x)=\dfrac{2t}{1+t^2}$ , but sure, we can derive this formula considering the double angle identity

$\sin(x)= 2\sin\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2}\right)$

Recall

$\sin \arctan t = \dfrac{t}{\sqrt{1 + t^2}} \text{ and } \cos \arctan t = \dfrac{1}{\sqrt{1 + t^2}}$

Thus,

$\sin(x)= 2 \left(\dfrac{t}{\sqrt{1 + t^2}}\right) \left(\dfrac{1}{\sqrt{1 + t^2}}\right) = \dfrac{2t}{1 + t^2}$

Similarly for cosine, consider the double angle identity

Thus,

$\cos(x)= \left(\dfrac{1}{\sqrt{1 + t^2}}\right)^2- \left(\dfrac{t}{\sqrt{1 + t^2}}\right)^2 = \dfrac{1}{t^2+1}-\dfrac{t^2}{t^2+1} =\dfrac{1-t^2}{1+t^2}$

Hence, we showed $\sin(x) \text { and } \cos(x)$

======================================================

$5\cos(x) =12\sin(x) +3, x \in [0, 2\pi ]$

Solving

$5\,\overbrace{\frac{1-t^2}{1+t^2}}^{\cos(x)} = 12\,\overbrace{\frac{2t}{1+t^2}}^{\sin(x)}+3$

$\implies \dfrac{5-5t^2}{1+t^2}= \dfrac{24t}{1+t^2}+3 \implies \dfrac{5-5t^2 -24t}{1+t^2}= 3$

$\implies 5-5t^2-24t=3\left(1+t^2\right) \implies -8t^2-24t+2=0$

$t = \dfrac{-(-24)\pm \sqrt{(-24)^2-4(-8)\cdot 2}}{2(-8)} = \dfrac{24\pm 8\sqrt{10}}{-16} = \dfrac{3\pm \sqrt{10}}{-2}$

$t=-\dfrac{3+\sqrt{10}}{2}\\t=\dfrac{\sqrt{10}-3}{2}$

Just note that

$\tan\left(\dfrac{x}{2}\right) = \dfrac{3\pm 8\sqrt{10}}{-2}$

and $\tan\left(\dfrac{x}{2}\right)$ is not defined for $x=k\pi , k\in\mathbb{Z}$

6 0

3 years ago

The number of violent crimes committed in a day possesses a distribution with a mean of 2.3 crimes per day and a standard deviat

marusya05 [52]

Answer:

B) approximately normal with mean 2.3 and standard deviation 2

Step-by-step explanation:

Given the following :

Mean crimes per day = 2.3

Standard deviation = 2

Number of samples = 100

Sampling distribution of the mean:

According to the central limit theorem:

Sample mean = population mean

2.3 = 2.3

The standard deviation of the sample (s) : ratio of the population standard deviation and square root of the sample size.

s = population standard deviation / √sample size

s = 2 / √100

s = 2 / 10

s = 0.2

The central limit theorem also posits that once ths sample size is large enough, the sampling of the sample mean will be approximately normal.

Hence, the distribution is approximately normal, with mean of 2.3 and standard deviation of 0.2

7 0

3 years ago