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3 years ago

Give the digits in the hundred thousands place and the ten thousands place. 175,384

Mathematics

1 answer:

Lady_Fox [76]3 years ago

3 0

Hundred-thousands: 1(00,000)

Ten-thousands: 7(0,000)

I'm not sure whether or not it's asking for the full number or just the digit, but I hope I helped!

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This famous clock tower has a clock face with a diameter of 23 feet. What is the approximate area of this clock face

pashok25 [27]

Answer:

Therefore the approximate area of this clock face is 415.3 feet².

Step-by-step explanation:

Given:

Clock is of Circular Shape,

Diameter = 23 feet

$Radius =r =\dfrac{diameter}{2}=\dfrac{23}{2}=11.5\ feet$

To Find :

Area of Clock Face = ?

Solution:

Area of Circle having Radius 'r' is given as,

$\textrm{Area of Circle}=\pi r^{2}$

Substituting the values we get

$\textrm{Area of Clock Face}=3.14\times (11.5)^{2}=415.265=415.3\ feet^{2}$

Therefore the approximate area of this clock face is 415.3 feet².

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3 years ago

A, 5+x = 17

Vlad [161]

A. 5+x=17, x = 17-5=12

B. x = 4 x 2.5 = 10

C. x = 27+12 = 39

D. x = 16:4 = 4

E. x = 9 : 3/4 = 12

Answer A and E

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2 years ago

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At a little-known vacation spot, taxi fares are a bargain. A 42 -mile taxi ride takes 49 minutes and costs $29.40. You want

dangina [55]

Answer:

about 13.50$

Step-by-step explanation:

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3 years ago

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Solve the equation m²+4m=-2

UNO [17]

Step 1: Subtract -2 from both sides.
m2+4m−(−2)=−2−(−2)
m2+4m+2=0
Step 2: Use quadratic formula with a=1, b=4, c=2.
m=−b±√b2−4ac2a
m=−(4)±√(4)2−4(1)(2)2(1)
m=−4±√82
m=−2+√2 or m=−2−√2

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3 years ago

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Coach Evans recorded the height, in inches of each player on his team. The results are shown.

marysya [2.9K]

Answer:

Step-by-step explanation:

Given:

Team heights (inches):

61, 57, 63, 62, 60, 64, 60, 62, 63

To find: IQRs (interquartile ranges) of the heights for the team

Solution:

A quartile divides the number of terms in the data into four more or less equal parts that is quarters.

For a set of data, a number for which 25% of the data is less than that number is known as the first quartile $(Q_1)$

For a set of data, a number for which 75% of the data is less than that number is known as the third quartile $(Q_3)$

Terms in arranged in ascending order:

$57,60,60,61,62,62,63,63,64$

Number of terms = 9

As number of terms is odd, exclude the middle term that is 62.

$Q_1$ is median of terms $57,60,60,61$

Number of terms (n) = 4

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{60+60}{2}=\frac{120}{2}=60$

So, $Q_1=60$

So, 25% of the heights of a team is less than 60 inches

$Q_3$ is the median of terms $62,63,63,64$

Median = $\frac{(\frac{n}{2})^{th} +(\frac{n}{2}+1)^{th} }{2} =\frac{2^{nd}+3^{rd}}{2} =\frac{63+63}{2}=\frac{126}{2}=63$

So, $Q_3=63$

So, 75% of the heights of a team is less than 63 inches

Interquartile range = $Q_3-Q_1=63-60=3$

The interquartile range is a measure of variability on dividing a data set into quartiles.

The interquartile range is the range of the middle 50% of the terms in the data.

So, 3 is the range of the middle 50% of the heights of the students.

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3 years ago