Given:
Sample Mean <span>= 30<span>
Sample size </span><span><span><span>= 1000</span></span><span>
</span></span></span>Population Standard deviation or <span><span><span>σ<span>=2</span></span><span>
</span></span>Confidence interval </span><span>= 95%</span>
to compute for the confidence interval
Population Mean or <span>μ<span><span>= sample mean ± (</span>z×<span>SE</span>)</span></span>
<span><span>where:</span></span>
<span><span>SE</span>→</span> Standard Error
<span><span>SE</span>=<span>σ<span>√n</span>= 30</span></span>√1000=0.9486
Critical Value of z for 95% confidence interval <span>=1.96</span>
<span>μ<span>=30±<span>(1.96×0.9486)</span></span><span>
</span></span><span>μ<span>=30±1.8594</span></span>
Upper Limit
<span>μ <span>= 30 + 1.8594 = 31.8594</span></span>
Lower Limit
<span>μ <span>= 30 − 1.8594 = <span>28.1406</span></span></span>
<span><span><span>
</span></span></span>
<span><span><span>answer: 28.1406<u<31.8594</span></span></span>
Carl is incorrect. Dave ate a higher fraction of snack bars, by 0.2 snack bars.
Carl had .5 left of a snack bar.
Dave had .3 left of a snack bar.
Tony had .5 left of a snack bar.
Gary had .0 left of a snack bar.
Tryone had .7 left of a snack bar.
If we add the above snack bars, there is a total of two remaining snack bars, meaning they only ate 12 of 14 snack bars.
Answer:
Accelerating to top speed, deaccelerating to finish line.
Step-by-step explanation:
If the runner kept a constant speed of 11 mph for the whole duration of his run (32 minutes), the distance he would have covered is:
This means that, in order to run the full 6.2 miles, the runner needs to reach a speed over 11 mph. Assume he starts from rest, while accelerating the runner reaches, and the surpasses, the 11 mph mark. Since his speed at the finish line is zero, the runner has to deaccelerate from his current running speed (which should be higher than 11 mph), passing through 11 mph and reaching zero at the finish line.
Answer:
see explanation
Step-by-step explanation:
Assuming you are factoring the expression
Given
4y² + 26y + 30 ← factor out 2 from each term
= 2(2y² + 13y + 15) ← factor the quadratic
Consider the factors of the product of the coefficient of the y² term and the constant term which sum to give the coefficient of the y- term.
product = 2 × 15 = 30 and sum = 13
the factors are 10 and 3
Use these factors to split the y- term
2y² + 10y + 3y + 15 ( factor the first/second and third/fourth terms )
= 2y(y + 5) + 3(y + 5) ← factor out (y + 5) from each term
= (y + 5)(2y + 3)
Thus
4y² + 26y + 30
= 2(y + 5)(2y + 3)
Answer:
- <em><u>5.6875 in</u></em>
Explanation:
At the point of tangency, the <em>tangent </em>to a circle and the <em>radius</em> form a right triangle (the radius is perpendicular to the tangent).
Here you are given the length of the tangent (6in), and the distance from the bisected vertex to the circle (2.75 in)
I tried to upload the drawing but the tool is not allowing it now.
In the figure:
- The length of the tangent (6 in) is one leg of the triangle
- The distance from vertex and the circle (2.75in) along with the radius forms the hypotenuse of the right triangle: 2.75 + r.
- The other leg is the radius, r.
Then, you can use Pythagorean theorem:
Solve:
- r² + 36 = r² + 5.5r + 7.5625
The solution is in inches: r = 5.6875 inches ← answer
