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nalin [4]
3 years ago
10

9.2\times 10^{-3}-1.3\times 10^{-3} 9.2×10 −3 −1.3×10 −3

Mathematics
2 answers:
g100num [7]3 years ago
8 0

Answer:

73

Step-by-step explanation:

9.2  x 10 - 3 - 1.3 x 10 - 3

92 -3 - 13- 3

73 is the answer if that the question

garik1379 [7]3 years ago
8 0

Answer:

(9.2 \times  {10}^{ - 3})  - (1.3 \times  {10}^{ - 3} ) \\  \\  = { \tt{(9.2 - 1.3) \times  {10}^{ - 3} }} \\  \\  = { \tt{7.9 \times  {10}^{ - 3} }}

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It rained 1 and 4/10 centimeters. You have to first figure out what adds up to 14, which is 1/10. Then you add 1 to get 15 and 3/10 to get 15 3/10.
3 0
3 years ago
a student played a computer game 500 times and won 370 of these game. he then won the next "x" games and lost none. he has now w
a_sh-v [17]

So, this is a proportion problem. If we express 75% as a fraction (3/4ths), we have:

\frac{370+x}{500+x}=\frac{3}{4}

Since he won all of his games, we add the number won (x), to the number previously won and divide it by the total games played which is just the previous games played plus the number of current games played, so since he lost no games we still add x to the denominator as well. Solving this gives us:

4(370+x)=3(500+x) \implies\\ 1480+4x=1500+3x \implies \\ x=20

That x value is 20.

3 0
3 years ago
Heather's class took a field trip to the science museum. They left school at 7:30 A.M. It took
vodka [1.7K]

Answer:

10:30 A.M

Step-by-step explanation:

5 0
3 years ago
Subject : Maths<br> Level : High school<br> Topic : Surds<br> Points : 72
Mila [183]

Answer:

  • a = 5

Step-by-step explanation:

<u>Simplify:</u>

  • √5(√8 + √18) =
  • √5*8 + √5*18 =
  • √40 + √90 =
  • √4*10 + √9*10 =
  • 2√10 + 3√10 =
  • (2 + 3)√10 =
  • 5√10

<u>The value of a:</u>

  • a√10 = 5√10
  • a = 5
6 0
3 years ago
A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard
bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44

In point a)  

\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}

  = \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2

value= (1- \frac{1}{k^2}) \times 100 \% =75\%

In point b)

\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma  \\\\= |(x-\mu)|\leq 2.5 \times 4.44  \\\\  = |(x-\mu)|\leq 1.11 \\\\ =  (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)

In point c)

\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)\\

In point d)

using standard normal variate

x=20.17\\\\  z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17

8 0
3 years ago
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