What is the greatest common factor of 36 and 25

A coin bank has 250 coins, dimes and quarters, worth $39.25. How many of each type of coin are there?

Brilliant_brown [7]

$x-\ number\ of\ dimes\\y-\ number\ of\ quarters\\\\ \left \{ {{x+y=250\ \ \ | x=250-y} \atop {0.1x+0.25y=39.25\ \ \ \ \ \ \ \ \ \ \ \ \ }} \right. \\\\ substitution\ method\\\\
0.1(250-y)+0.25y=39.25\\\\
25-0.1y+0.25y=39.25\\\\
25+0.15y=39.25\ \ \ \ | subtract\ 25\\\\0.15y=14.25\ \ \ \ | divide\ by\ 0.15\\\\y=95\\\\x=250-95=
155\\\\
There\ are\ 155\ dimes\ and\ 95\ quarters.$

8 0

3 years ago

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 30,393 miles, with a standard

Pie

Answer:

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

Step-by-step explanation:

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

Normal probability distribution:

Problems of normally distributed samples are 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 $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 30393, \sigma = 2876, n = 37, s = \frac{2876}{\sqrt{37}} = 472.81$

What is the probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

This probability is the pvalue of Z when X = 30393 + 339 = 30732 subtracted by the pvalue of Z when X = 30393 - 339 = 30054. So

X = 30732

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

By the Central Limit Theorem

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

$Z = \frac{30732 - 30393}{472.81}$

$Z = 0.72$

$Z = 0.72$ has a pvalue of 0.7642.

X = 30054

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

$Z = \frac{30054 - 30393}{472.81}$

$Z = -0.72$

$Z = -0.72$ has a pvalue of 0.2358

0.7642 - 0.2358 = 0.5284

52.84% probability that the sample mean would differ from the population mean by less than 339 miles in a sample of 37 tires if the manager is correct

4 0

3 years ago

An elephant can run of a mile in 36 seconds. At this rate, which expression can be used to determine how fast an elephant runs i

Mice21 [21]

Answer:

The expression that can be used to determine how fast an elephant runs in miles per hour is 100 miles per hour

Solution:

Given that,

Distance covered by an elephant in 36 seconds is equal to 1 mile

Therefore, distance covered by the elephant in 1 second is $\frac{1}{36}$ mile

Therefore, distance covered by the elephant in 60 seconds (60 seconds = 1 minute) is

$\frac{1}{36} \times 60=\frac{5}{3} \text { miles }$

Therefore, distance covered by the elephant in 60 minute (60 minute = 1 hour) is

$\frac{5}{3} \times 60=100 \text { miles per hour }$

Hence the expression that can be used to determine how fast an elephant runs in miles per hour is 100 miles per hour

7 0

3 years ago

I'm confused on this (this is 12 year old math)

Gemiola [76]

Answer: the top box can be the linear equation y=40x+15 and the bottom box is 40

Step-by-step explanation:

If you swim 15 more minutes than your friend and your swam 55 minutes, that means your friend only swam 40 minutes. The set amount of time that your friend swims for x is 40 minutes. So the additional time you swim is b, 15 minutes.

7 0

2 years ago

On a lunch counter, there are 3 oranges, 5 apples, and 2 bananas. If 3 pieces of fruit are selected, find the probability that 1

muminat

Answer:

$Probability = \frac{1}{24}$