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gulaghasi [49]
4 years ago
13

PLEASE HELP!!!! WILL GET MAX POINTS

Mathematics
2 answers:
solmaris [256]4 years ago
8 0
3.71 is the answer. 52 divided by 14 = 3.71

kvv77 [185]4 years ago
3 0
To find the likelihood, you need to find decimal under 1 because it's a probability question. In probability, the value should be between 0 and 1, with 1 meaning a 100% chance that event will happen.

Since you're looking at someone who only wears mediums, you can ignore the rest of the chart. There are 52 people total surveyed that wore mediums. Look for where yellow and medium intersect. 14 of the people that wore mediums also wanted a yellow shirt.

The probability of finding a student that wears a medium that wants a yellow shirt is then 14/52 = 0.27.
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vladimir2022 [97]
A , because if you multiple you have to distribute
3 0
3 years ago
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Which is the equation in slope-intercept form of the line that contains points E and F?
lys-0071 [83]

Answer:

  • C. y = 4x - 12

Step-by-step explanation:

<u>The coordinates of E and F are:</u>

  • E(4, 4) and F(2, -4)

<u>Find the slope:</u>

  • m = (- 4 - 4)/(2 - 4) = - 8 / - 2 = 4

<u>Use point-slope form and point E:</u>

  • y - 4 = 4(x - 4)

<u>Convert into slope-intercept form:</u>

  • y = 4x - 16 + 4
  • y = 4x - 12

The matching choice is C

8 0
2 years ago
Read 2 more answers
Help calculus module 8 DBQ<br><br> please show work
igor_vitrenko [27]

1. The four subintervals are [0, 2], [2, 3], [3, 7], and [7, 8]. We construct trapezoids with "heights" equal to the lengths of each subinterval - 2, 1, 4, and 1, respectively - and the average of the corresponding "bases" equal to the average of the values of R(t) at the endpoints of each subinterval. The sum is then

\dfrac{R(0)+R(2)}2(2-0)+\dfrac{R(2)+R(3)}2(3-2)+\dfrac{R(3)+R(7)}2(7-3)+\dfrac{R(7)+R(8)}2(7-8)=\boxed{24.83}

which is measured in units of gallons, hence representing the amount of water that flows into the tank.

2. Since R is differentiable, the mean value theorem holds on any subinterval of its domain. Then for any interval [a,b], it guarantees the existence of some c\in(a,b) such that

\dfrac{R(b)-R(a)}{b-a)=R'(c)

Computing the difference quotient over each subinterval above gives values of 0.275, 0.3, 0.3, and 0.26. But just because these values are non-zero doesn't guarantee that there is definitely no such c for which R'(c)=0. I would chalk this up to not having enough information.

3. R(t) gives the rate of water flow, and R(t)\approx W(t), so that the average rate of water flow over [0, 8] is the average value of W(t), given by the integral

R_{\rm avg}=\displaystyle\frac1{8-0}\int_0^8\ln(t^2+7)\,\mathrm dt

If doing this by hand, you can integrate by parts, setting

u=\ln(t^2+7)\implies\mathrm du=\dfrac{2t}{t^2+7}\,\mathrm dt

\mathrm dv=\mathrm dt\implies v=t

R_{\rm avg}=\displaystyle\frac18\left(t\ln(t^2+7)\bigg|_{t=0}^{t=8}-\int_0^8\frac{2t^2}{t^2+7}\,\mathrm dt\right)

For the remaining integral, consider the trigonometric substitution t=\sqrt 7\tan s, so that \mathrm dt=\sqrt 7\sec^2s\,\mathrm ds. Then

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\frac{7\tan^2s}{7\tan^2s+7}\sec^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}\tan^2s\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\int_0^{\tan^{-1}(8/\sqrt7)}(\sec^2s-1)\,\mathrm ds

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan s-s\right)\bigg|_{s=0}^{s=\tan^{-1}(8/\sqrt7)}

R_{\rm avg}=\displaystyle\ln71-\frac{\sqrt7}4\left(\tan\left(\tan^{-1}\frac8{\sqrt7}\right)-\tan^{-1}\frac8{\sqrt7}\right)

\boxed{R_{\rm avg}=\displaystyle\ln71-2+\frac{\sqrt7}4\tan^{-1}\frac8{\sqrt7}}

or approximately 3.0904, measured in gallons per hour (because this is the average value of R).

4. By the fundamental theorem of calculus,

g'(x)=f(x)

and g(x) is increasing whenever g'(x)=f(x)>0. This happens over the interval (-2, 3), since f(x)=3 on [-2, 0), and -x+3>0 on [0, 3).

5. First, by additivity of the definite integral,

\displaystyle\int_{-2}^xf(t)\,\mathrm dt=\int_{-2}^0f(t)\,\mathrm dt+\int_0^xf(t)\,\mathrm dt

Over the interval [-2, 0), we have f(x)=3, and over the interval [0, 6], f(x)=-x+3. So the integral above is

\displaystyle\int_{-2}^03\,\mathrm dt+\int_0^x(-t+3)\,\mathrm dt=3t\bigg|_{t=-2}^{t=0}+\left(-\dfrac{t^2}2+3t\right)\bigg|_{t=0}^{t=x}=\boxed{6+3x-\dfrac{x^2}2}

6 0
3 years ago
What is the third quartile of this set?<br> 165, 85, 105, 152, 144, 140, 160, 80
Oxana [17]
It’s the 5 and 6 numbers 152 and 144
4 0
4 years ago
Read 2 more answers
Mark estimates that the distance from his house to his school is 15.5 mi. The actual distance is 14.5 mi.
lapo4ka [179]
So the error is 1 mile

I ask what % of 15.5 = 1

How you write this;

x/100 *15.5 = 1

15.5x = 100

x = 100/15.5

x = 6.451629 round up 6.45

% error is 6.45 %
7 0
3 years ago
Read 2 more answers
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