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3 years ago

Please help with number one!! thx

Mathematics

1 answer:

agasfer [191]3 years ago

5 0

5032.34
Y = 0.82 * 6137
Y = 5032.34

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The data below are the ages and systolic blood pressures (measured in millimeters of mercury) of 9 randomly selected adults. Wha

seraphim [82]

Answer:

$\sum_{i=1}^n x_i =459$

$\sum_{i=1}^n y_i =1227$

$\sum_{i=1}^n x^2_i =24059$

$\sum_{i=1}^n y^2_i =168843$

$\sum_{i=1}^n x_i y_i =63544$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=24059-\frac{459^2}{9}=650$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=63544-\frac{459*1227}{9}=967$

And the slope would be:

$m=\frac{967}{650}=1.488$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{459}{9}=51$

$\bar y= \frac{\sum y_i}{n}=\frac{1227}{9}=136.33$

And we can find the intercept using this:

$b=\bar y -m \bar x=136.33-(1.488*51)=60.442$

So the line would be given by:

$y=1.488 x +60.442$

And then the best predicted value of y for x = 41 is:

$y=1.488*41 +60.442 =121.45$

Step-by-step explanation:

For this case we assume the following dataset given:

x: 38,41,45,48,51,53,57,61,65

y: 116,120,123,131,142,145,148,150,152

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$

So we can find the sums like this:

$\sum_{i=1}^n x_i =459$

$\sum_{i=1}^n y_i =1227$

$\sum_{i=1}^n x^2_i =24059$

$\sum_{i=1}^n y^2_i =168843$

$\sum_{i=1}^n x_i y_i =63544$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=24059-\frac{459^2}{9}=650$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}=63544-\frac{459*1227}{9}=967$

And the slope would be:

$m=\frac{967}{650}=1.488$

Nowe we can find the means for x and y like this:

$\bar x= \frac{\sum x_i}{n}=\frac{459}{9}=51$

$\bar y= \frac{\sum y_i}{n}=\frac{1227}{9}=136.33$

And we can find the intercept using this:

$b=\bar y -m \bar x=136.33-(1.488*51)=60.442$

So the line would be given by:

$y=1.488 x +60.442$

And then the best predicted value of y for x = 41 is:

$y=1.488*41 +60.442 =121.45$

3 0

3 years ago

Brett is hosting a party for 16 guests,including himself.Brett orders 2 meatball,1 turkey,1 veggie party-size sandwiches. Each s

schepotkina [342]

Answer:

Step-by-step explanation:

Given:

Brett orders 2 meatball,1 turkey,1 veggie party-size sandwiches.

Each sandwich costs the same amount.

The total amount cost of the sandwiches, including tax, is $54.36.

Question asked:

What is the cost of each sandwich including tax?

Solution:

Let cost of each sandwich including tax = $x$

Total number of sandwich = 2 + 1+ 1 = 4

As each sandwich costs the same amount and total amount of cost of 4 sandwiches, including tax, is $54.36, the equation will be:-

$x+x+x+x=\$54.36\\4x=\$54.36$

By dividing both sides by 4,

$x=\$13.59$

Thus, the cost of each sandwich including tax is $13.59

6 0

3 years ago

Please help me out. What are these ??

Nata [24]

Invert at one point

6 0

3 years ago

SIMPLYFY THE EXPRESSION WRITE THE ANSWER AS A POWER (5^4)^3

Anastaziya [24]

Answer:

5^12

Step-by-step explanation:

You multiply both roots

5 0

2 years ago

Assume the random variable X is normally distributed with mean mu equals 50 and standard deviation sigma equals 7. Compute the p

defon

Answer:

$P(X>34) = 0.9889$

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 50

Standard Deviation, σ = 7

We are given that the distribution of random variable X is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X greater than 34)

$P( X > 34) = P( z > \displaystyle\frac{34 - 50}{7}) = P(z > -2.2857)$

$= 1 - P(z \leq -2.2857)$

Calculation the value from standard normal z table, we have,

$P(X>34) = 1 - 0.0111= 0.9889= 98.89\%$

The attached image shows the normal curve.

4 0

3 years ago