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2 years ago

14

A caterer charges a setup fee of $50, plus $20 per person. How much will the caterer charge if 35 people attend the party, and t

he customer has a coupon for $100 off the total?

1 answer:

tiny-mole [99]2 years ago

3 0

Answer:

$650

Step-by-step explanation:

35 people x $20 per person = $700

$700 + $50 set up fee = $750 dollars total

$750 - $100 coupon = $650

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Find the tangent line equation of the curve at the given point. Y=arcsin(7x) at the point where x=sqrt2/14

Mumz [18]

Answer:

$Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

Step-by-step explanation:

The equation of the curve is

$Y = sin^{-1}(7x)$

To find the equation of tangent we need to differentiate this equation w.r.t x

So, differentiating we get

$Y'=\frac{7}{\sqrt{1-49x^2} }$

This would give the slope of the tangent line at any given point of which x coordinate is known. In the present case it is $x = \sqrt{\frac{1}{7} }$

Then slope would accordingly be

$Y'=\frac{7}{\sqrt{1-49/49} }$

= ∞

For, $x = \sqrt{\frac{1}{7} }$ , $Y = sin^{-1}(7/7)= \pi/2$

Equation of tangent line, in the point slope form, would be $Y-\frac{\pi}{2} =\frac{\pi}{2} (x+\sqrt{\frac{1}{7}})$

4 0

2 years ago

Solving, H = $<br> and C = $

lapo4ka [179]

Answer:

h=1.29 c=0.79

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

The weights of adult male birds of a certain species are normally distributed with a mean of 27.5 grams and a standard deviation

Korvikt [17]

Answer:

The z-score of a male bird of this species with a weight of 29.37 grams is 1.7.

Step-by-step explanation:

We are given that the weights of adult male birds of a certain species are normally distributed with a mean of 27.5 grams and a standard deviation of 1.1 grams.

<u><em>Let X = weights of adult male birds of a certain species</em></u>

So, X ~ Normal( $\mu=27.5,\sigma^{2} =1.1^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{ X-\mu}{\sigma } }$ ~ N(0,1)

where, $\mu$ = population mean weight = 27.5 grams

$\sigma$ = standard deviation = 1.1 grams

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

SO, the z-score of a male bird of this species with a weight of 29.37 grams is given by;

<u>Z score</u> = $\frac{ X-\mu}{\sigma } }$ = $\frac{ 29.37-27.5}{1.1 } }$

= <u>1.7</u>

4 0

2 years ago

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of days an

solniwko [45]

Answer:

(a) 283 days

(b) 248 days

Step-by-step explanation:

The complete question is:

The pregnancy length in days for a population of new mothers can be approximated by a normal distribution with a mean of 268 days and a standard deviation of 12 days. (a) What is the minimum pregnancy length that can be in the top 11% of pregnancy lengths? (b) What is the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths?

Solution:

The random variable <em>X</em> can be defined as the pregnancy length in days.

Then, from the provided information $X\sim N(\mu=268, \sigma^{2}=12^{2})$ .

(a)

The minimum pregnancy length that can be in the top 11% of pregnancy lengths implies that:

P (X > x) = 0.11

⇒ P (Z > z) = 0.11

⇒ <em>z</em> = 1.23

Compute the value of <em>x</em> as follows:

$z=\frac{x-\mu}{\sigma}\\\\1.23=\frac{x-268}{12}\\\\x=268+(12\times 1.23)\\\\x=282.76\\\\x\approx 283$

Thus, the minimum pregnancy length that can be in the top 11% of pregnancy lengths is 283 days.

(b)

The maximum pregnancy length that can be in the bottom 5% of pregnancy lengths implies that:

P (X < x) = 0.05

⇒ P (Z < z) = 0.05

⇒ <em>z</em> = -1.645

Compute the value of <em>x</em> as follows:

$z=\frac{x-\mu}{\sigma}\\\\-1.645=\frac{x-268}{12}\\\\x=268-(12\times 1.645)\\\\x=248.26\\\\x\approx 248$

Thus, the maximum pregnancy length that can be in the bottom 5% of pregnancy lengths is 248 days.

8 0

2 years ago

Ashely is making $500 a week at her job. She receives a 20% pay raise. A year later she takes a 5% pay cut. How much money does

zheka24 [161]

Answer:

<em>$570 a week</em>

Step-by-step explanation:

<u>Percentage Calculations </u>

One tricky fact about percentages is that they are usually performed over the last computed quantity. It makes them non-additive. For example, a raise of 10% followed by a 10% some time later, is NOT a 20% of total raise

The question states Ashely is making $500 a week at her job and that she receives a 20% pay raise. After that raise she will make

$\$500*(1+20/100)=\$500(1.2)=\$600 \text{ a week}$

A year later she takes a 5% pay cut. We must be careful to take that percentage from the last earnings. She now makes

$\$600(1-5/100)=\$600(0.95)=\$570\ a\ week$

3 0

3 years ago