You have 4 framed portraits. How many ways can you arrange all 4 of them on the top of your bookshelf?.

What is the standard form of the equation of the circle in the graph?

Dmitriy789 [7]

C(1;2)
a = 1
b = 2
r = 4

(x-a)² + (y-b)² = r²

(x-1)² + (y-2)² = 4²

(x-1)² + (y-2)² = 16

5 0

3 years ago

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Only some of the students in the Level 1 swim lessons move up to Level 2 at the end of

larisa86 [58]

The list that orders the fractions from least to greatest is:

$\frac{7}{12}, \frac{2}{3}, \frac{17}{24}, \frac{3}{4}$

<h3>Ordering of fractions</h3>

To order the fractions from least to greatest, we have to realize their conversion to decimal.

To realize the conversion, the numerator is divided by the denominator, as ordering decimal numbers are easier than fractions.

Hence, the fractions converted to decimal are:

$\frac{2}{3} = 0.6667$ .

$\frac{7}{12} = 0.5833$ .

$\frac{17}{24} = 0.7083$

$\frac{3}{4} = 0.75$

Hence, the list from least to greatest is, according to the results of the conversions:

$\frac{7}{12}, \frac{2}{3}, \frac{17}{24}, \frac{3}{4}$

To learn more about ordering of fractions, you can take a look at brainly.com/question/26139168

7 0

2 years ago

These triangles are congruent by which triangle congruence postulate ?

svp [43]

(a) SSS (side side side)
every side on one triangle is correspondingly equal to that of the same side in the other triangle

3 0

2 years ago

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One warehouse had 185 tons of coal, another one had 237 tons. The first warehouse delivered 15 tons of coal to its clients daily

sdas [7]

Given

One warehouse had 185 tons of coal, another one had 237 tons

The first warehouse delivered 15 tons of coal to its clients daily, the second one delivered 18 tons of coal to its clients daily.

Find out in how many days will the second warehouse have 1.5 times more coal than the first one.

To proof

Let us assume that the number of days of delivered the coal = x

Take two cases

First case

As given in the question

One warehouse had coal = 185 tons

The first warehouse delivered 15 tons of coal to its clients daily

(As first warehouse delivered 15 tons of coal to its clients daily than the total amount of coal decrease daily as days increase)

Than the equation becomes for first warehouse

= ( 185 - 15x)

( this equation represent the amount of coal left after x days)

Second case

second warehouse had coal = 237 tons

the second one delivered 18 tons of coal to its clients daily.

(As first warehouse delivered 18 tons of coal to its clients daily than the total amount of coal decrease daily as days increase)

Than the equation becomes for the second warehouse

= ( 237 - 18x)

( this equation represent the amount of coal left after x days)

As given in the question

the second warehouse have 1.5 times more coal than the first one

Thus

( 237 -18x ) = 1.5× ( 185 - 15x)

237 - 18x = 277.5 - 22.5x

22.5x - 18x = 277.5 - 237

4.5x = 40.5

x = 9

Therefore the second warehouse have 1.5 times more coal than the first one in 9 days.

Hence proved

7 0

3 years ago

The distribution of SAT II Math scores is approximately normal with mean 660 and standard deviation 90. The probability that 100

gayaneshka [121]

Using the <em>normal distribution and the central limit theorem</em>, it is found that there is a 0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

<h3>Normal Probability Distribution</h3>

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

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

It 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, which is the percentile of X.

By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

In this problem:

The mean is of 660, hence $\mu = 660$ .

The standard deviation is of 90, hence $\sigma = 90$ .

A sample of 100 is taken, hence $n = 100, s = \frac{90}{\sqrt{100}} = 9$ .

The probability that 100 randomly selected students will have a mean SAT II Math score greater than 670 is <u>1 subtracted by the p-value of Z when X = 670</u>, hence:

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

By the Central Limit Theorem

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

$Z = \frac{670 - 660}{9}$

$Z = 1.11$

$Z = 1.11$ has a p-value of 0.8665.

1 - 0.8665 = 0.1335.

0.1335 = 13.35% probability that 100 randomly selected students will have a mean SAT II Math score greater than 670.

To learn more about the <em>normal distribution and the central limit theorem</em>, you can take a look at brainly.com/question/24663213

7 0

1 year ago