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
Vesna [10]
3 years ago
9

What is the smallest three-digit number divisible by the first three prime numbers and the first three composite numbers?

Mathematics
1 answer:
Flura [38]3 years ago
5 0
Its 120. Divisible by 1,2,3,4,6,8.
You might be interested in
The mean life of a television set is 119119 months with a standard deviation of 1414 months. If a sample of 7474 televisions is
Pepsi [2]

Answer:

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean

In this problem, we have that:

\mu = 119, \sigma = 14, n = 74, s = \frac{14}{\sqrt{74}} = 1.6275

What is the probability that the sample mean would differ from the true mean by less than 1.11 months?

This is the pvalue of Z when X = 119 + 1.1 = 120.1 subtracted by the pvalue of Z when X = 119 - 1.1 = 117.9. So

X = 120.1

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{120.1 - 119}{1.6275}

Z = 0.68

Z = 0.68 has a pvalue of 0.7517

X = 117.9

Z = \frac{X - \mu}{s}

Z = \frac{117.9 - 119}{1.6275}

Z = -0.68

Z = -0.68 has a pvalue of 0.2483

0.7517 - 0.2483 = 0.5034

0.5034 = 50.34% probability that the sample mean would differ from the true mean by less than 1.1 months

3 0
3 years ago
Derek flips a coin and rolls a number cube at the same time. Derek rolls an even number. Find the probability that the coin came
Andreas93 [3]
Coin: 50%; 1/2 chance to get heads
dice: 17%; 1/6 chance to get 2
7 0
3 years ago
Read 2 more answers
Give answer fast...<br>it's very important for me....<br>then I give you 20 thanks...(◍•ᴗ•◍)❤​
Vedmedyk [2.9K]

Answer:

I have the answer

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What is the distance between the points (10, 10) and (4, 2)
ElenaW [278]
How does it want the answer to be? Like a fraction? If so, the answer would be 8/6 since rise over run. 8 up and 6 to the left.
7 0
3 years ago
What are the minimum, first quartile, median, third quartile, and maximum of the data set? 2, 13, 17, 14, 9, 3, 16, 12
Alex777 [14]

So before anything, rearrange the data so that it's in ascending order: {2,3,9,12,13,14,16,17}

Next, the minimum and maximum are as they seem: the smallest and largest numbers in the data. Looking at our data, <u>2 is our minimum and 17 is our maximum.</u>

Next, to find the median, find the number that is in the middle of the data set. If there isn't one number in the middle, find the average of the two middle numbers to get your median:

\{2,3,9,\overbrace{\boxed{12,13,}}^{\textsf{middle numbers}}14,16,17\}\\\\\frac{12+13}{2}\\\\\frac{25}{2}\\\\12.5

<u>The median is 12.5.</u>

Next, to find the first quartile, or the lower quartile, find the "median" of the numbers to the left of the median.

\overbrace{\{2,\boxed{3,9,}12,}^{\textsf{left of median}}13,14,16,17\}\\\\\frac{3+9}{2}\\\\\frac{12}{2}\\\\6

<u>The first quartile is 6.</u>

Next, to find the third quartile, or the upper quartile, find the "median" of the numbers to the right of the median.

\{2,3,9,12,\overbrace{13,\boxed{14,16,}17\}}^{\textsf{right of median}}\\\\\frac{14+16}{2}\\\\\frac{30}{2}\\\\15

<u>The third quartile is 15.</u>

3 0
3 years ago
Other questions:
  • A hiker at the top of a mountain spots his car back in the parking lot at the
    8·1 answer
  • I need help solving this, and one other question, please!
    5·2 answers
  • Can some one friend me plz
    8·1 answer
  • Use graphing technology to find the range of the function f(x)=x2 +6x+9
    8·1 answer
  • What is the value of f(–1)?
    13·1 answer
  • Complete 6,7 for 5-8 points.
    5·2 answers
  • Do someone one want to help me with this ?
    15·1 answer
  • F is a function that describes a sequence and is therefore defined over the positive
    11·1 answer
  • Plz i will give u 93 point and a brainlylist and I will add u as a friend
    9·2 answers
  • The equations of three lines are given below.
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!