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nordsb [41]
3 years ago
9

Georgia has 9 meters of ribbon and wants to make 80 spirit ribbons that are each 12 cm long. Does she have enough ribbon?

Mathematics
2 answers:
avanturin [10]3 years ago
6 0
Yes, she has more than enough ribbon
r-ruslan [8.4K]3 years ago
5 0
No, she will be 60cm short
there are 100cm in a meter
she will have 900cm of ribbon but making 80 ribbons of 12cm would require 960cm
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Slav-nsk [51]
All you need to do is divide .88 by 6.8.

.88/6.8= 0.129 (round upt to .13 cents)

I hope this helps!
3 0
3 years ago
Read 2 more answers
a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-
Tresset [83]

Answer:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

M = \frac{1}{2}(Lower + Upper)

Where

Lower \to Lower class interval

Upper \to Upper class interval

So, we have:

Class 63-65:

M = \frac{1}{2}(63 + 65) = 64

Class 66 - 68:

M = \frac{1}{2}(66 + 68) = 67

When the computation is completed, the frequency distribution will be:

\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1}  & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7}  & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

\bar x = 70.2903

\sigma = 3.5795

<em>See attachment for result of 1-VarStats</em>

8 0
3 years ago
A woman is randomly selected from the 18-24 age group. For women of this group, systolic blood pressures (in mm Hg) are normally
dybincka [34]

Answer:

0.0274

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If X\sim N(114.8,\ 13.1), then Z\sim N(0,1)

and

Pr(X>140)=Pr(Z>1.9237).

Use table for normal distribution probabilities to get that

Pr(Z>1.9237)=1-Pr(Z\le 1.9237)=1-0.9726=0.0274.

7 0
3 years ago
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Lemur [1.5K]

Answer:

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multiply them all

5 0
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What is the mean absolute deviation and what does it tell you about data set
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It is the average distance between each data value and the mean<span>.</span>
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