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Delicious77 [7]
3 years ago
13

Which of the following is used to name a segment that has an endpoint at point A and another endpoint at point B?

Mathematics
1 answer:
MakcuM [25]3 years ago
4 0

You would write AB with a horizontal line segment over the letters A and B

So you would write this \overline{AB}

If you cannot draw that on your computer, then I would stick to saying "segment AB".

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The lowest recorded temperature in Wisconsin is -55°F on February 4, 1996. Use the expression 5(F-32)/9 to find this temperature
Lorico [155]

Answer:

-48.3° Celsuis

Step-by-step explanation:

Mark as brainllest

4 0
3 years ago
Vitamin D, whether ingested as a dietary supplement or produced naturally when sunlight falls on the skin, is essential for stro
tigry1 [53]

Answer:

a) The 98% confidence interval would be given (0.182;0.218).

b) We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

c) If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

d) Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

Step-by-step explanation:

Data given and notation  

n=2700 represent the random sample taken    

X represent the people in England who are deficient in Vitamin D

\hat p=0.2 estimated proportion of England people who are deficient in Vitamin D

\alpha=0.02 represent the significance level

Confidence =0.98 or 98%

z would represent the statistic (variable of interest)    

p= population proportion of England people who are deficient in Vitamin D

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 98% confidence interval the value of \alpha=1-0.98=0.02 and \alpha/2=0.01, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=2.33

And replacing into the confidence interval formula we got:

0.20 - 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.182

0.20 + 2.33 \sqrt{\frac{0.2(1-0.2)}{2700}}=0.218

And the 98% confidence interval would be given (0.182;0.218).

Part b

We are 98% confident that the true proportion of of England people who are deficient in Vitamin D is between 0.182 and 0.218.

Part c

If repeated samples were taken and the 98% confidence interval computed for each sample, 98% of the intervals would contain the population proportion.

Part d

Yes since the confidence interval not contains the value 0.25, we can refute the claim at 2% of significance.

3 0
3 years ago
30 points What is the distance between points B and C on the coordinate plane shown below?
Galina-37 [17]

Answer:

looks like 9.5, but kinda hard to tell

3 0
3 years ago
If a=4 and b=5, what is the value of the following expression? 3 (a+3b)-4
gavmur [86]

Answer: The answer is 23

Step-by-step explanation:


8 0
3 years ago
Read 2 more answers
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.

5 0
4 years ago
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