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3 years ago

Can you find 10/21 of 3/ 10

Mathematics

1 answer:

hoa [83]3 years ago

5 0

<span>0.1428571428 would be the correct answer</span>

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You have $5.00 in quarters and dimes. You have a total of 41 coins. How many of the coins are quarters?

Dmitry_Shevchenko [17]

One because if you put two then the would be 50 and thats over 41

8 0

3 years ago

Simplify the expression below. (2n/6n+4)(3n+2/3n-2)

andrezito [222]

Answer:

$\frac{n}{3n-2}$

Step-by-step explanation:

Given the expression:

$\frac{2n}{(6n+4)} \cdot \frac{3n+2}{3n-2}$

Using the distributive property: $a \cdot (b+c) =a\cdot b+a\cdot c$

$\frac{2n}{2(3n+2)} \cdot \frac{3n+2}{3n-2}$

Simplify:

$\frac{2n}{2} \cdot \frac{1}{3n-2}$

Simplify:

$\frac{n}{3n-2}$

Therefore, the simplified given expression is, $\frac{n}{3n-2}$

7 0

3 years ago

Read 2 more answers

12% is 90% of what number?

Andrews [41]

12% = 0.12

90% = 0.9

$0.12 = 0.9 \times x\\\\x = \frac{0.12}{0.9}=\boxed{1.33333}$

7 0

3 years ago

A triangle has two sides of length 20 and 11. What is the largest possible whole-number length for the third side?

charle [14.2K]

Answer:

Step-by-step explanation:

20+11=31

It can't be exactly 31 but it can be any number less than 31, so the nearest whole number down is 30

8 0

2 years ago

The sugar content of the syrup is canned peaches is normally distributed. Assumethe can is designed to have standard deviation 5

andreyandreev [35.5K]

Answer: 0.50477

Step-by-step explanation:

Given : The sugar content of the syrup is canned peaches is normally distributed.

We assume the can is designed to have standard deviation $\sigma=5$ milligrams.

The sampling distribution of the sample variance is chi-square distribution.

Also,The data yields a sample standard deviation of $s=4.8$ milligrams.

Sample size : n= 10

Test statistic for chi-square : $\chi^2=\dfrac{s^2(n-1)}{\sigma^2}$

i.e. $\chi^2=\dfrac{(4.8)^2(10-1)}{(5)^2}=8.2944$

Now, P-value = $P(\chi^2>8.2944)=0.50477$ [By using the chi-square distribution table for p-values.]

Hence, the chance of observing the sample standard deviation greater than 4.8 milligrams = 0.50477

8 0

2 years ago