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exis [7]
3 years ago
15

If the scale of the map is 1 inch = 125 miles, two points that are 3.75 inches apart will actually be approximately how many mil

es away from each other? 292 375 469 1,125
Mathematics
1 answer:
lys-0071 [83]3 years ago
3 0

Answer:

469 miles

Step-by-step explanation:

1 in = 125 miles

3.75 in =?

3.75 x 125 = 468.75 miles

468.75 rounds up to 469 miles

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Find the rule for 2, 1, 0, -1
natita [175]
It’s 2-1=1
1-1= 0
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3 years ago
Solve for q.<br><br> s = 4p + 4q solve the equation
Harlamova29_29 [7]

Answer:

q = \frac{s -4p}{4}\\

Step-by-step explanation:

s = 4p +4q \\ 4p +4q = s \\ 4p +4q -4p = s -4p \\ 4q = s - 4p \\ \frac{4q}{4} = \frac{s -4p}{4} \\ q = \frac{s -4p}{4}

4 0
3 years ago
PLEASE HELP ME FIRST GETS BRAINLIEST!
Fudgin [204]
Measures of variation: <span> ways to describe measures of central tendency for a data.
<em>
</em><em>the most common measure of variation is range, so in this case, the smallest number is 1, and the largest is 17. This means the range is 16.
</em>
Interquartile Range:

<em>Lower q: {1, 2, 4}
</em><em>Upper q: {8, 12, 17}
</em>Here you have to subtract 12 from 2, which gives you 10.

Mean absolute deviation:
Alright, let's first find the mean...

(1+2+4+5+8+12+17)
---------------------------= 7
                7

What is the distance of each number from 7?

1: 6 units
2: 5 units
4: 3 units
5: 2 units
8: 1 unit
12: 5 units
17: 10 units

After adding all of these together, and dividing by 7, you will get...

<em>4.571428 </em></span>≈ <em>4.6 or 5.
</em>
<span>
I hope these answers help you! Please give me brainliest!
</span>
5 0
3 years ago
Each of 100 students in the Allen School can only take 1 CSE class each, between the four classes CSE 311, CSE 312, CSE 331, and
Naddika [18.5K]

Answer:

P(a=31,b=39,c=7,d=23) = 0.000668

Step-by-step explanation:

Sample space, n = 100

Let the number of students signed up for CSE 311 = a

Let the number of students signed up for CSE 312 = b

Let the number of students signed up for CSE 331 = c

Let the number of students signed up for CSE 332 = d

Probability of taking CSE 311, P_a = 0.3

Probability of taking CSE 312, P_b = 0.4

Probability of taking CSE 331, P_c = 0.1

Probability of taking CSE 332, P_d = 0.2

P(a,b,c,d) = \frac{n!}{a! b! c! d!} p_a^{a}  p_b^{b}  p_c^{c}  p_d^{d} \\P(a=31,b=39,c=7,d=23) = \frac{100!}{31! 39! 7! 23!} *  0.3^{31} * 0.4^{39} * 0.1^{7}  0.2^{23}\\P(a=31,b=39,c=7,d=23) = \frac{4.58*10^{111}}{2.13*10^{56}* 5040 }* (1.57*10^{-55})\\P(a=31,b=39,c=7,d=23) = 0.000668

3 0
3 years ago
Which describes the first step in solving this equation? 1/2x = 40
Lisa [10]
Cross multiplication.
5 0
3 years ago
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