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2 years ago

How many inches are in 2miles?

Mathematics

2 answers:

noname [10]2 years ago

4 0

Answer:

2 miles = 126720 (looked it up)

Step-by-step explanation:

abruzzese [7]2 years ago

4 0

There are 126,720 inches in 2 miles

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-2c + 3c - 5 - 4c + 7

kaheart [24]

Answer:−

3c+2

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

⦁ Find the sum of the series . 15 sigma n=1, (2n-1) Show your work

sergejj [24]

Answer:

225

Step-by-step explanation:

The sum of this series is given by the formula:

$S_n=\frac{n}{2}[2a+(n-1)d]$

Where

S_n is the sum

a is the first term (we get this by plugging in n = 1 into "2n-1", we have 2(1) - 1 = 1)

n is the number of terms (the sum is defined for n = 1 to n = 15, so 15 terms)

d is the common different (the difference is successive terms. Here, 2nd term would be 2(2) - 1 = 3, and first term was 1, so d = 3 -1 = 2)

plugging in the info, we will get the sum:

$S_n=\frac{n}{2}[2a+(n-1)d]\\S_{15}=\frac{15}{2}[2(1)+(15-1)(2)]\\=225$

The sum is 225

4 0

2 years ago

Emery borrowed money from her brother to buy a new phone, and is paying off a fixed amount each week. After 2 weeks, she will ow

jarptica [38.1K]

Answer: A. The original price was $608. B. $76 each week C. Quite useful, as it keeps track of the amount of money she still needs to owe each week.

Step-by-step explanation: For part A, 5 week subtracted by 2 weeks is 3 weeks. 456-228 is 228. Then divide 228 by 3 and you get $76. Then do 2x76 to get 152 and add it to 456, and you get 608.

5 0

2 years ago

Choose true or false.

just olya [345]

Answer:

True

Step-by-step explanation:

7 0

3 years ago

Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot

SashulF [63]

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

$P(x)=^nC_xp^x(1-p)^{n-x}$

The required probability = $P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473$

Hence, the required probability = 0.0473

5 0

2 years ago