Ask question

Ask question

All categories

3 years ago

8

During one month there were 7 days of precipitation.What if there had only been 3 days of precipitation that month?How would tha

t change the measures of center?

1 answer:

faust18 [17]3 years ago

7 0

We've hit on a case where a measure<span> of </span>center does<span> not provide all the information spread or variability </span>there<span> is </span>in month-to-month precipitation<span>. based on how busy each </span>month has been in<span> the past, lets managers plan Unit 6: Standard Deviation | Student Guide | Page </span>3<span> </span>If<span> you sum the deviations from the mean, (. ).</span>

You might be interested in

Help me please i’m really confused on it :) thankyou

Nata [24]

A is 6:55

B is 37 Minutes

C is 15:12

Step-by-step explanation:

so basically its just adding minutes to a clock, like if its 1:00 o clock and 40 minutes pass it will be 1:40 or if its 12:00 and 25 minutes pass it will be 12:55. There's 60 minutes in an hour so if the time is, let's say, 12:45 and 30 minutes go by itd be 1:15.

C is a little more tricky but thats just because its military time.:) basically instead of going to 12 am to 1 pm it would go from 12:00 AM to 13:00 PM

5 0

2 years ago

How many solutions are there to the equation 10= a + b + c with each of a, b, and c a whole number greater than or equal to zero

djverab [1.8K]

If <em>a</em> is fixed and <em>b</em>,<em>c</em> are unknowns then the equation <em>b</em>+<em>c</em>=10-<em>a</em> has 11-<em>a</em> solutions. They are pairs (b,c): (0,10-a), (1,9-a), (2,8-a), ... (10-a,0). As <em>a</em> runs from 0 to 10 we have total number of solutions (11-0)+(11-1)+...(11-1)=11+10+...+1=(1+11)*11/2=66.

4 0

3 years ago

Directions: Calculate the area of a circle using 3.14x the radius

Leokris [45]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

As we know ~

Area of the circle is :

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

And radius (r) = diameter (d) ÷ 2

[ radius of the circle = half the measure of diameter ]

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

<h3>Problem 1</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 4.4\div 2$

$\qquad \sf \dashrightarrow \:r = 2.2 \: mm$

Now find the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(2.2)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {4.84}^{}$

$\qquad \sf \dashrightarrow \:area \approx 15.2 \: \: mm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 2</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 3.7 \div 2$

$\qquad \sf \dashrightarrow \:r = 1.85 \: \: cm$

Bow, calculate the Area ~

$\qquad \sf \dashrightarrow \: \pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (1.85) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 3.4225 {}^{}$

$\qquad \sf \dashrightarrow \:area \approx 10.75 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 3 </h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (8.3) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 68.89$

$\qquad \sf \dashrightarrow \:area \approx216.31 \: \: cm {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>Problem 4</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 5.8 \div 2$

$\qquad \sf \dashrightarrow \:r = 2.9 \: \: yd$

now, let's calculate area ~

$\qquad \sf \dashrightarrow \:3.14 \times {(2.9)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 8.41$

$\qquad \sf \dashrightarrow \:area \approx26.41 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 5</h3>

$\qquad \sf \dashrightarrow \:r = d \div 2$

$\qquad \sf \dashrightarrow \:r = 1 \div 2$

$\qquad \sf \dashrightarrow \:r = 0.5 \: \: yd$

Now, let's calculate area ~

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times (0.5) {}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 0.25$

$\qquad \sf \dashrightarrow \:area \approx0.785 \: \: yd {}^{2}$

・．━━━━━━━†━━━━━━━━━．・

<h3>problem 6</h3>

$\qquad \sf \dashrightarrow \:\pi {r}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times {(8)}^{2}$

$\qquad \sf \dashrightarrow \:3.14 \times 64$

$\qquad \sf \dashrightarrow \:area = 200.96 \: \: yd {}^{2}$

➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖➖

8 0

2 years ago

What is the range of 149 111 171 166 152 129 162 144

valentinak56 [21]

Answer:

range = bigger number-smaller number

biggest number=166

smallest number=111

166-111

=55

8 0

3 years ago

Read 2 more answers

Suppose Eric and Isaac both drive the same car, and have the same

Leto [7]

Answer:

eric

Step-by-step explanation:

just took the test

5 0

3 years ago

Read 2 more answers