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sammy [17]
3 years ago
10

The difference between the most hours of sleep and the least of these students was 2.5 hours. True or false?​

Mathematics
1 answer:
SashulF [63]3 years ago
8 0

The correct anwer is False

Explanation

According to the graph, it can be seen that a student has four hours of sleep as the minimum number of hours of sleep; two students have six hours of sleep, four students have six and a half hours of sleep, four have seven hours of sleep, three have seven and a half hours of sleep, five have eight hours of sleep, and one has eight and a half hours of sleep as maximum hours of sleep. Therefore, it can be affirmed that the statement that the difference between the maximum amount and the minimum number of hours is two and a half hours is false because between four hours and eight and a half hours there are four and a half hours of difference. So, the correct answer is False.

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30 points!!<br> What is the sum of the first six terms of the series?<br> 48 - 12 + 3 - 0.75 +...
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The sum of the first six terms is 38.39

Step-by-step explanation:

This is a geometric sequence since the common difference between each term is -\frac{1}{4}

Thus, r=-\frac{1}{4}

To find the sum of first six terms, we need to find the fifth and sixth term of the sequence.

To find the fifth term:

The general form of geometric sequence is a_{n}=a_{1} \cdot r^{n-1}

To find the fifth term, substitute n=5 in a_{n}=a_{1} \cdot r^{n-1}

\begin{aligned}a_{5} &=(48) \cdot\left(-\frac{1}{4}\right)^{5-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{4} \\&=(48)\left(\frac{1}{256}\right) \\a_{5} &=0.1875\end{aligned}

To find the sixth term, substitute n=6 in a_{n}=a_{1} \cdot r^{n-1}

\begin{aligned}a_{6} &=(48) \cdot\left(-\frac{1}{4}\right)^{6-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{5} \\&=(48)\left(-\frac{1}{1024}\right) \\a_{5} &=-0.046875\end{aligned}

To find the sum of the first six terms:

The general formula to find Sn for |r| is S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}

\begin{aligned}S_{6} &=\frac{48\left(1-\left(-\frac{1}{4}\right)^{6}\right)}{1-\left(-\frac{1}{4}\right)} \\&=\frac{48\left(1-\frac{1}{4096}\right)}{1+\frac{1}{4096}} \\&=\frac{48(0.95)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{47.9904}{5} \\&=38.39\end{aligned}

Thus, the sum of first six terms is 38.39

5 0
3 years ago
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