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3 years ago

Phillip wanted to leave a 15% tip. Phillip's bill was $36.00. How much will he leave for a tip?

Mathematics

2 answers:

Anastaziya [24]3 years ago

4 0

Your answer would be letter B. $5.40

stealth61 [152]3 years ago

4 0

B 5.40 would be his tip because you multiply 36 by .15

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Which equations represent the line that is perpendicular to the line 5x − 2y = −6 and passes through the point (5, −4)? Select t

nignag [31]

For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

On the other hand we have that if two lines are perpendicular, then the product of their slopes is -1. So:

$m_ {1} * m_ {2} = - 1$

The given line is:

$5x-2y = -6\\-2y = -6-5x\\2y = 5x + 6\\y = \frac {5} {2} x + \frac {6} {2}\\y = \frac {5} {2} x + 3$

So we have:

$m_ {1} = \frac {5} {2}$

We find $m_ {2}:$ $m_ {2} = \frac {-1} {\frac {5} {2}}\\m = - \frac {2} {5}$

So, a line perpendicular to the one given is of the form:

$y = - \frac {2} {5} x + b$

We substitute the given point to find "b":

$-4 = - \frac {2} {5} (5) + b\\-4 = -2 + b\\-4 + 2 = b\\b = -2$

Finally we have:

$y = - \frac {2} {5} x-2$

In point-slope form we have:

$y - (- 4) = - \frac {2} {5} (x-5)\\y + 4 = - \frac {2} {5} (x-5)$

ANswer:

$y = - \frac {2} {5} x-2\\y + 4 = - \frac {2} {5} (x-5)$

3 0

3 years ago

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132 beads is split into 96 beads and 36 beads. in simplest form , what is the ratio of the split

mars1129 [50]

Answer:

16:6

Step-by-step explanation:

96:36

Highest common factor is 6

divide both sides by 6

16:6

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3 years ago

HELP PLEASE 10 POINTS!!!

7nadin3 [17]

Answer:

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2 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

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3 years ago

A certain bowling league has weekly matches where two teams compete against each other. Each team has four players whose individ

hram777 [196]

Answer: 30 Pins

Step-by-step explanation:

I solved it on khan academy

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2 years ago

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