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Afina-wow [57]
3 years ago
8

Khan academy : Sasha has been saving the money she earns from babysitting for several months. She has a total of $96.75 dollar s

aved. Round the amount of money to the nearest ten.
Mathematics
1 answer:
AVprozaik [17]3 years ago
6 0

Answer:

$100.00

Step-by-step explanation:

If we are rounding to the nearest 10, then we only have to look at 96. Since 6 is bigger than 5, we round up to 100.

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4c+5


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amm1812

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yes

Step-by-step explanation:

it is a right triangle

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What is the value of tan60.cot60​
sladkih [1.3K]

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1

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6 0
2 years ago
*HELP PLEASE*
aksik [14]
If two events are disjoint (i.e. mutually exclusive) they cannot both occur and so we can determine probabilties by simple addition:

P(AorB)=P(A)+P(B) 
P(AorB)=\frac{40}{61}+\frac{2}{61} 
P(AorB)=\frac{42}{61} 
P(AorB)\approx 68.8\% 

A more complete answer would use a joint probability table but that's difficult to type out/format correctly here. 
7 0
3 years ago
A student scored 75 and 97 on her first two quizzes. Write and solve a compound inequality to find the possible values for a thi
Rainbow [258]

Answer:

85 \le \frac{75 + 97 + n}{3} \le 90; 83 \le n \le 98

Step-by-step explanation:

Score of the first two quizzes: 75 and 97

Let n represent the third score

Average score would be: \frac{75 + 97 + n}{3}.

Given that the average score fall between 85 and 90, this can be represented by the compound inequality as shown below:

85 \le \frac{75 + 97 + n}{3} \le 90

Solve for n in each statement that makes up the compound inequality:

85 \le \frac{75 + 97 + n}{3}

85 \le \frac{172 + n}{3}

Multiply both sides by 3

85*3 \le 172 + n

255 \le 172 + n

Subtract 172 from each side of the inequality

255 - 172 \le n

83 \le n

Also,

\frac{75 + 97 + n}{3} \le 90

\frac{172 + n}{3} \le 90

Multiply both sides by 3

172 + n \le 90*3

172 + n \le 270

Subtract 172 from both sides of the inequality

n \le 270 - 172

n \le 98

Combining both together, the possible values of her third quiz score would be:

83 \le n \le 98

3 0
2 years ago
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