1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tom [10]
2 years ago
11

140.3 is 23% of what number?

Mathematics
2 answers:
LenaWriter [7]2 years ago
6 0

600

explanation: well, since 140.3 divided by 600 is equal to 0.233833333... means that is that 140.3 is the 23 percent of 600

Illusion [34]2 years ago
3 0

Answer:

610

Step-by-step explanation:

\frac{140.3}{0.23} =610

You might be interested in
5 times the sum of 40 and 80. what is the numerical expression that would represent the phrase?
vlabodo [156]
<h2>Answer:</h2>

<u>The expression is</u><u> 5(40+80)</u>

<h2>Step-by-step explanation:</h2>

The sum of 40 and 80 can be written as

(40+80)

and when we take the 5 times of the sum

we can write as

5(40+80)

3 0
3 years ago
One angle of an isosceles triangle measures 138°. Which other angles could be in that isosceles triangle?
BartSMP [9]

Answer:

2 21 degree angles

Step-by-step explanation:

An isosoles triangle has two congruent angles and every triangle's sum of interior angles equals 180 degrees. 138*2>180, so the 138 degree angle cannot be congruent to any of the other angles. Therefore:

180=138+2x

42=2x

21 degrees=x

3 0
3 years ago
WILL MARK BRAINLIEST
nordsb [41]
For# 1 the answer would be
-3/5

for #2 the answer would be
3/4

i hope this helps :)
7 0
3 years ago
What is the measure of ∠DEG?<br> A) 133° <br> B) 103° <br> C) 77° <br> D) 67°
nordsb [41]
Hmmmmmm
Let me think for a second

3 0
3 years ago
Read 2 more answers
Is the confidence interval affected by the fact that the data appear to be from a population that is not normally​ distributed?
sp2606 [1]

Answer:

D. No, because the sample size is large enough.

Step-by-step explanation:

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

If the sample size is higher than 30, on this case the answer would be:

D. No, because the sample size is large enough.

And the reason is given by The Central Limit Theorem since states if the individual distribution is normal then the sampling distribution for the sample mean is also normal.

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

If the sample size it's not large enough n<30, on that case the distribution would be not normal.

7 0
3 years ago
Other questions:
  • Corey measure a crayon box writhing his paper clip ruler. about how long is the box
    13·1 answer
  • I need help I'm dumb
    9·1 answer
  • Please help asap please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
    14·1 answer
  • Describe the points with a double inequality. I need the answer quick. Thanks!
    5·1 answer
  • What is the equation of the line that passes through the point (6,14) and is parallel to the line with the following equation? y
    13·1 answer
  • PLS HELP ME ASAL FOR ALL 6!!! (MUST SHOW WORK FOR ALL!!) + LOTS OF POINTS!!
    12·1 answer
  • What is the slope if the line that passes through the pair of points (-5.5, 6.1) (-2.5, 3.1)
    9·1 answer
  • What is the image of (-8, 2) after a reflection over the y-axis?
    10·1 answer
  • Help me i really need this .no spam​
    8·1 answer
  • How do i solve this?
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!