Answer:
2m^4 - 11m³ + m + 8
Step-by-step explanation:
(2m - 5m³) - (5m + 3m³ - 7m^4) + (8 - 3m³ - 5m^4 + 4m)
2m - 5m³ - 5m - 3m³ + 7m^4 + 8 - 3m³ - 5m^4 + 4m
7m^4 - 5m^4 - 5m³ - 3m³ - 3m³ + 2m - 5m + 4m + 8
2m^4 - 11m³ + m + 8
To solve this problem you must follow the proccedure shown below:
1. Amount of lemonade in pints:
1 <span>quart of water=2 pints
1 pint of lemon
(9 ounces of honey)(0.0625 pints/1 ounce)=0.56 pints
Total in pints=2 pints+1 pint+0.56 pints
Total in pints=3.56 pints
2. </span>Amount of lemonade in<span> cups:
Total in cups=(3.56 pints)(2 cups/1 pint)
Total in cups=7.12 cups
3. </span>Amount of lemonade in<span> ounces:
Total in ounces=(7.12 cups)(8 ounces/1 cup)
Total in ounces=56.96 ounces</span>
Answer:
7. is 2 squared. I didn't show my work but I am very sure that's the answer.
9514 1404 393
Answer:
1. ∠EDF = 104°
2. arc FG = 201°
3. ∠T = 60°
Step-by-step explanation:
There are a couple of angle relationships that are applicable to these problems.
- the angle where chords meet is half the sum of the measures of the intercepted arcs
- the angle where secants meet is half the difference of the measures of the intercepted arcs
The first of these applies to the first two problems.
1. ∠EDF = 1/2(arc EF + arc UG)
∠EDF = 1/2(147° +61°) = 1/2(208°)
∠EDF = 104°
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2. ∠FHG = 1/2(arc FG + arc ES)
128° = 1/2(arc FG +55°) . . . substitute given information
256° = arc FG +55° . . . . . . multiply by 2
201° = arc FG . . . . . . . . . subtract 55°
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3. For the purpose of this problem, a tangent is a special case of a secant in which both intersection points with the circle are the same point. The relation for secants still applies.
∠T = 1/2(arc FS -arc US)
∠T = 1/2(170° -50°) = 1/2(120°)
∠T = 60°
Answer:
a) Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
b) If the true mean is 190 days, Type II error can be made.
Step-by-step explanation:
Let mu be the mean life of the batteries of the company when it is used in a wireless mouse
Null and alternative hypotheses are:
: mu=183 days
: mu>183 days
Type II error happens if we fail to reject the null hypothesis, when actually the alternative hypothesis is true.
That is if we conclude that mean life of the batteries of the company when it is used in a wireless mouse is at most 183 days, but actually mean life is 190 hours, we make a Type II error.
