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3 years ago

Sophia has read 75 pages in a 250-page book. What percent of the book has Sophia read?

Mathematics

2 answers:

Cerrena [4.2K]3 years ago

6 0

Sophia has read 30% of the book

yaroslaw [1]3 years ago

4 0

If Sophia has read 75 pages and the book is 250
you take 75 and divide it by 250
your answer will be .3
witch would be 30% as a percentage

hope i helped Dosvedanya

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Help please confused

Sindrei [870]

Answer:

00.5 is a answer A IS A ANSWER

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2 years ago

Renee is simplifying the expression (7) (StartFraction 13 over 29 EndFraction) (StartFraction 1 over 7 EndFraction). She recogni

dsp73

Answer:

The correct option is commutative property.

Step-by-step explanation:

The expression that Renee is simplifying is:

$(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})$

It is provided that, Renee recognizes that 7 and $\frac{1}{7}$ are reciprocals, so she would like to find their product before she multiplies by $\frac{13}{29}$ .

The associative property of multiplication states that:

$a\times b\times c=(a\times b)\times c=a\times (b\times c)$

The commutative property of multiplication states that:

$a\times b\times c=a\times c\times b=c\times a\times b$

The distributive property of multiplication states that:

$a\cdot (b+c)=a\cdot b+a\cdot c$

The identity property of multiplication states that:

$a\times 1=a\\b\times 1=b$

So, Renee should use the commutative property of multiplication to find the product of 7 and $\frac{1}{7}$ ,

$(7)\cdot(\frac{13}{29})\cdot(\frac{1}{7})=(7\times\frac{1}{7})\times\frac{13}{29}=\frac{13}{29}$

Thus, the correct option is commutative property.

5 0

3 years ago

Read 2 more answers

Solve the following equations: (a) x^11=13 mod 35 (b) x^5=3 mod 64

tino4ka555 [31]

$x^{11}=13\pmod{35}\implies\begin{cases}x^{11}\equiv13\equiv3\pmod5\\x^{11}\equiv13\equiv6\pmod7\end{cases}$

By Fermat's little theorem, we have

$x^{11}\equiv (x^5)^2x\equiv x^3\equiv3\pmod5$

$x^{11}\equiv x^7x^4\equiv x^5\equiv6\pmod 7$

5 and 7 are both prime, so $\varphi(5)=4$ and $\varphi(7)=6$ . By Euler's theorem, we get

$x^4\equiv1\pmod5\implies x\equiv3^{-1}\equiv2\pmod5$

$x^6\equiv1\pmod7\impleis x\equiv6^{-1}\equiv6\pmod7$

Now we can use the Chinese remainder theorem to solve for $x$ . Start with

$x=2\cdot7+5\cdot6$

Taken mod 5, the second term vanishes and $14\equiv4\pmod5$ . Multiply by the inverse of 4 mod 5 (4), then by 2.

$x=2\cdot7\cdot4\cdot2+5\cdot6$

Taken mod 7, the first term vanishes and $30\equiv2\pmod7$ . Multiply by the inverse of 2 mod 7 (4), then by 6.

$x=2\cdot7\cdot4\cdot2+5\cdot6\cdot4\cdot6$

$\implies x\equiv832\pmod{5\cdot7}\implies\boxed{x\equiv27\pmod{35}}$

$x^5\equiv3\pmod{64}$

We have $\varphi(64)=32$ , so by Euler's theorem,

$x^{32}\equiv1\pmod{64}$

Now, raising both sides of the original congruence to the power of 6 gives

$x^{30}\equiv3^6\equiv729\equiv25\pmod{64}$

Then multiplying both sides by $x^2$ gives

$x^{32}\equiv25x^2\equiv1\pmod{64}$

so that $x^2$ is the inverse of 25 mod 64. To find this inverse, solve for $y$ in $25y\equiv1\pmod{64}$ . Using the Euclidean algorithm, we have

64 = 2*25 + 14

25 = 1*14 + 11

14 = 1*11 + 3

11 = 3*3 + 2

3 = 1*2 + 1

=> 1 = 9*64 - 23*25

so that $(-23)\cdot25\equiv1\pmod{64}\implies y=25^{-1}\equiv-23\equiv41\pmod{64}$ .

So we know

$25x^2\equiv1\pmod{64}\implies x^2\equiv41\pmod{64}$

Squaring both sides of this gives

$x^4\equiv1681\equiv17\pmod{64}$

and multiplying both sides by $x$ tells us

$x^5\equiv17x\equiv3\pmod{64}$

Use the Euclidean algorithm to solve for $x$ .

64 = 3*17 + 13

17 = 1*13 + 4

13 = 3*4 + 1

=> 1 = 4*64 - 15*17

so that $(-15)\cdot17\equiv1\pmod{64}\implies17^{-1}\equiv-15\equiv49\pmod{64}$ , and so $x\equiv147\pmod{64}\implies\boxed{x\equiv19\pmod{64}}$

5 0

2 years ago

the national flag of scotland has a white diagonal on a blue backround Tracy is sewing a scottish flag she knows that the length

dybincka [34]

Answer:

12 feet

Step-by-step explanation:

By Pythagoras theorem, a²+b²=c², where c represents the diagonal and a and b represents the other sides of a right angled triangle

As a flag is a rectangle, the two triangles cut would be right angled triangles

Therefore, a²+9²=15²

a²+81=225

a²=144

a=√144

a=12 feet

7 0

2 years ago

Suppose that the population mean for income is $50,000, while the population standard deviation is 25,000. If we select a random

Fudgin [204]

Answer:

Probability that the sample will have a mean that is greater than $52,000 is 0.0057.

Step-by-step explanation:

We are given that the population mean for income is $50,000, while the population standard deviation is 25,000.

We select a random sample of 1,000 people.

Let $\bar X$ = sample mean

The z-score probability distribution for sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = $50,000

$\sigma$ = population standard deviation = $25,000

n = sample of people = 1,000

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the sample will have a mean that is greater than $52,000 is given by = P( $\bar X$ > $52,000)

P( $\bar X$ > $52,000) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{52,000-50,000}{\frac{25,000}{\sqrt{1,000} } }$ ) = P(Z > 2.53) = 1 - P(Z $\leq$ 2.53)

= 1 - 0.9943 = 0.0057

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.53 in the z table which has an area of 0.9943.

Therefore, probability that the sample will have a mean that is greater than $52,000 is 0.0057.

5 0

2 years ago