Given:
difference in the mean weight gain is 0.60 grams
standard deviation of the difference in sample mean is 0.305
68% confidence interval for the population mean difference is a) 0.305
0.60 <u>+</u> 1 * 0.305
0.60 + 0.305 = 0.905
0.60 - 0.305 = 0.295
95% confidence interval for the population mean difference is c) 0.61
0.60 <u>+</u> 2 * 0.305
0.60 + 0.61 = 1.21
0.60 - 0.61 = -0.01
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Answer:
In order to have ran 33 miles, Bobby would have to attend <em>32 track practices.</em>
Step-by-step explanation:
Solving this problem entails of uncovering the amount of track practices Bobby must attend in order to have ran 33 miles. Start by reading the problem carefully to break down the information provided.
You can see that Bobby has already ran one mile on his own. This is important to remember for later. The problem also states that he expects to run one mile at every track practice.
Setting up an equation will help us solve. Here is how we could set up the equation:
(<em>amount of miles already ran</em> = 1) + (<em>number of track practices</em> = x) = (<em>total miles to run</em> = 33)
1 + x = 33
The equation is now in place. You can solve this, or isolate <em>'x',</em> by using the subtraction property of equality. This means we will subtract one from both sides of the equation, thus isolating the variable.
1 + x = 33
1 - 1 + x = 33 - 1
x = 32
The variable is the only term left on the left side of the equation. This means Bobby must attend track practice <em>32 times</em> in order to have ran 33 miles.
Answer:yes
Step-by-step explanation:
Answer:
-56/9
Step-by-step explanation:
By Vieta's formulas,
$r + s = -\frac{4}{3}$ and $rs = \frac{12}{3} = 4.$ Squaring the equation $r + s = -\frac{4}{3},$ we get
$r^2 + 2rs + s^2 = \frac{16}{9}.$ Therefore,
$r^2 + s^2 = \frac{16}{9} - 2rs = \frac{16}{9} - 2 \cdot 4 = -\frac{56}{9}}$
