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Olin [163]
3 years ago
6

A chocolate company makes dark chocolate bars. The quality control department of the company has found that the weights of the c

hocolate bars follow a normal distribution, with a mean of 85 grams and standard deviation of 1.2 grams. To ensure a consistent product, the company has a policy to reject any chocolate bars weighing less than 82.6 grams or more than 87.4 grams. Approximately what percent of chocolate bars are rejected?
a. 0.12%
b. 0.15%
c. 0.30%
d. 2.50%
e. 5.00%
Mathematics
2 answers:
Aleksandr [31]3 years ago
8 0
I believe that it is D.
devlian [24]3 years ago
3 0
D and e i’m pretty sure :)
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3 years ago
I need help with this I don't understand this math
Dmitriy789 [7]

The exponential regression equation is y = 172.21(0.99)ˣ and at x = 60 minutes y = 94.22 degree Fahrenheit and at x = 240 minutes y = 15.43 degree Fahrenheit.

<h3>What is correlation?</h3>

It is defined as the relation between two variables which is a quantitative type and gives an idea about the direction of these two variables.

\rm r = \dfrac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{{[n\sum x^2- (\sum x)^2]}}\sqrt{[n\sum y^2- (\sum y)^2]}}

We have data shown in the table:

We can find the exponential regression equation using the data shown.

\rm y = ab^x

After calculating from the data:

a = 172.21 and b = 0.99

\rm y = 172.21(0.99)^x

The correlation coefficient will be:

r = -0.99

The range will be all real numbers.

The domain will be y > 0

As we can see in the graph x ∈R and y>0

After an hour,

Plug x = 60 minutes in the exponential regression equation:

\rm y = 172.21(0.99)^{60}

y = 94.22 degree Fahrenheit

After 4 hours, plug x = 240

\rm y = 172.21(0.99)^{240}

y = 15.43 degree Fahrenheit

 

Similarly, we can find any data from the exponential regression equation.

Thus, the exponential regression equation is y = 172.21(0.99)ˣ and at x = 60 minutes y = 94.22 degree Fahrenheit and at x = 240 minutes y = 15.43 degree Fahrenheit.

Learn more about the correlation here:

brainly.com/question/11705632

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7 0
2 years ago
Find the six trig function values of the angle 240*Show all work, do not use calculator
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Solution:

Given:

240^0

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

sin240^0=sin(180+60)

Using the trigonometric identity;

sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny

Hence,

\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

cos240^0=cos(180+60)

Using the trigonometric identity;

cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny

Hence,

\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\  \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\  \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

tan240^0=tan(180+60)

Using the trigonometric identity;

tan(180+x)=tan\text{ }x

Hence,

\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\  \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}

To get cosec 240 degrees:

\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\  \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}

To get sec 240 degrees:

\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\  \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\  \\ Thus, \\ sec240^0=-2 \end{gathered}

To get cot 240 degrees:

\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\  \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\  \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\  \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}

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An educated guess or a estimation


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