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3 years ago

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hen entering large numbers in the answer box, do not use commas. For example, enter 1276400‎ for the number 1,276,400. Do not en

ter 1,276,400‎. If you accidentally enter commas, you will receive feedback as a reminder. Answer the following question by typing the numeric answer into the answer box. What is the sum of 9260 and 3240?

1 answer:

Gelneren [198K]3 years ago

3 0

Answer:

12500

Step-by-step explanation:

Sum is addition.Add the first number to the second one

9260

<u>+</u><u>3240</u>

<u> 12500 </u>

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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 box which measures 70cm X 36cm X 12cm is to be covered by a canvas. How many meters of canvas of width 80cm would be require

grigory [225]

Answer:

142.2 meters.

Step-by-step explanation:

We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.

Let us find total surface area of box using surface area formula of cuboid.

$\text{Total surface area of cuboid}=2(lb+bh+hl)$ , where,

$l$ = Length of cuboid,

$b$ = Breadth of cuboid,

$w$ = Width of cuboid.

$\text{Total surface area of box}=2(70\cdot36+36\cdot 12+12\cdot 70)$

$\text{Total surface area of box}=2(2520+432+840)$

$\text{Total surface area of box}=2(3792)$

$\text{Total surface area of box}=7584$

Therefore, the total surface area of box will be 7584 square cm.

To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.

$\text{Width of canvas* Length of canvass}=\text{Total surface area of 150 boxes}$

$80\text{ cm}\times\text{ Length of canvass}=150\times 7584\text{cm}^2$

$\text{ Length of canvass}=\frac{150\times 7584\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=\frac{1137600\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=14220\text{ cm}$

Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100}\times\frac{m}{cm}$

$\text{ Length of canvass}=142.20\text{ m}$

Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.

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3 years ago

The owner of a small pet supply store wants to open a second store in another city, but he only wants to do so if more than one-

Andrei [34K]

Answer:

The answer is "The null hypothesis was rejected".

Step-by-step explanation:

Following are the right-tailed test:

Calculating the null and alternative hypothesis:

$H_0 : p = 0.3333\\\\H_a : p > 0.3333\\\\n = 150\\\\x = 64\\\\\hat{p} = \frac{x}{n} = \frac{64}{150} = 0.4267\\\\P_0 = 0.3333\\\\1 - P_0 = 1 - 0.3333 = 0.6667\\\\$

$z = \frac{\hat{p} - P_0}{[\sqrt{\frac{P_0 \times (1 - P_0 )}{n}}]}$

$= \frac{0.4267 - 0.3333}{[\sqrt{\frac{(0.3333 \times 0.6667)}{150}}]}\\\\= 2.426$

Calculating the right-tailed test:

$P(z > 2.426) = 1 - P(z < 2.426) = 1 - 0.9924 = 0.0076\\\\P-value = 0.0076\\\\\alpha = 0.05\\\\P-value < \alpha$

Therefore, we reject the null hypothesis.

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2 years ago

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kow [346]

12inches because 3 is pi and 2 is Radius so 3x2squared=12

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3 years ago

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3 years ago