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lukranit [14]
3 years ago
14

Ava's birthday cake is about 11 7/10 cm long. her candles stick up from the cake an additional 5 4/10 cm. what was the total hig

ht of the cake and the candles?{ the answer is 17.1}
Mathematics
2 answers:
tamaranim1 [39]3 years ago
4 0
17.1 is the correct answer
Virty [35]3 years ago
3 0

Total height of the cake and the candles is 17.1 cm

Step-by-step explanation:

  • Step 1: Convert the mixed fractions into decimal expressions.

11 7/10 cm = 117/10 = 11.7 cm

5 4/10 cm = 54/10 = 5.4 cm

  • Step 2: Find total height of the cake and the candles

Total height = 11.7 + 5.4 = 17.1 cm

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2 years ago
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Calculate the sample mean and sample variance for the following frequency distribution of heart rates for a sample of American a
spin [16.1K]

Answer:

Mean = 68.9

s^2 =18.1 --- Variance

Step-by-step explanation:

Given

\begin{array}{cccccc}{Class} & {51-58} & {59-66} & {67-74} & {75-82} & {83-90} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

Solving (a): Calculate the mean.

The given data is a grouped data. So, first we calculate the class midpoint (x)

For 51 - 58.

x = \frac{1}{2}(51+58) = \frac{1}{2}(109) = 54.5

For 59 - 66

x = \frac{1}{2}(59+66) = \frac{1}{2}(125) = 62.5

For 67 - 74

x = \frac{1}{2}(67+74) = \frac{1}{2}(141) = 70.5

For 75 - 82

x = \frac{1}{2}(75+82) = \frac{1}{2}(157) = 78.5

For 83 - 90

x = \frac{1}{2}(83+90) = \frac{1}{2}(173) = 86.5

So, the table becomes:

\begin{array}{cccccc}{x} & {54.5} & {62.5} & {70.5} & {78.5} & {86.5} \ \\ {Frequency} & {6} & {3} & {11} & {13} & {4} \ \end{array}

The mean is then calculated as:

Mean = \frac{\sum fx}{\sum f}

Mean = \frac{54.5*4+62.5*3+70.5*11+78.5*13+86.5*4}{6+3+11+13+4}

Mean = \frac{2547.5}{37}

Mean = 68.9 -- approximated

Solving (b): The sample variance:

This is calculated as:

s^2 =\frac{\sum (x - \overline x)^2}{\sum f - 1}

So, we have:

s^2 =\frac{(54.5-68.9)^2+(62.5-68.9)^2+(70.5-68.9)^2+(78.5-68.9)^2+(86.5-68.9)^2}{37 - 1}

s^2 =\frac{652.8}{36}

s^2 =18.1 -- approximated

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2 years ago
The area of polygon MNOPQR = Area of a rectangle that is 9 square units + Area of a rectangle that is ___ square units. (Input w
vekshin1
You have the polygon MNOPQR which can be expressed as two rectangles pasted one next to each other.

To see the two rectangles in the picture, you can draw a line parallel to segment MR througn point N.

From the original picture you can state the dimensions of both rectangles.

Call S, the point where the line that you drew intercepts the segment RQ.

Then one rectangle is MNSR and the other rectangle is OPQS.

The measures of the sides of the rectangle MNSR are:

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So its area is base * height, which you can all A1.

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- segment PQ = segment OS = height

So its area is base * height, which you can call A2.

Then the area of the polygon MNOPQRS is A1 + A2. One of them is 9 u^2 and the other is what the answer is asking for, and that you have calculated above.

With this procedure you can tell the value needed.   

 
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