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3 years ago

856 divided by 57 somebody please help?

Mathematics

2 answers:

fgiga [73]3 years ago

7 0

Answer:

15.0175439

Step-by-step explanation:

seropon [69]3 years ago

4 0

856/57 = 15.017

Please tell me if it is correct! :)

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Given sec theta=-4/3 and 90

KATRIN_1 [288]

What is ur question?

8 0

3 years ago

The graph below shows the solution to which system of inequalities?

nlexa [21]

Answer:

Option D.

Step-by-step explanation:

Consider option D. $y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3$

Take point (0,0)

On putting this point in inequation $y\leq \frac{3x}{4}+10$ , we get

$0\leq 10$ which is true . So, solution is region towards the origin i,e region below the line $y= \frac{3x}{4}+10$ including the line itself .

On putting (0,0) in inequation $y\leq \frac{-x}{2}-3$ , we get $0\leq -3$ which is false , so solution is region away from the origin i.e region below line $y= \frac{-x}{2}-3$ including the line itself .

So, common solution to both the inequations is the shaded part in the given figure .

In other words, we can say that the graph shown in the given figure represents system of equations: $y\leq \frac{3x}{4}+10\,,\,y\leq \frac{-x}{2}-3$

5 0

3 years ago

Recent census data indicated that 14.2% of adults between the ages of 25 and 34 live with their parents. A random sample of 125

kicyunya [14]

Answer:

The probability is $P(14 < X < 20 ) = 0.5354$

Step-by-step explanation:

From the question we are told that

The proportion that live with their parents is $\r p = 0.142$

The sample size is n = 125

Given that there are two possible outcomes and that this outcomes are independent of each other then we can say the Recent census data follows a Binomial distribution

i.e

$X \ \~ \ B( \mu , \sigma )$

Now the mean is evaluated as

$\mu = n * \r p$

$\mu = 125 * 0.142$

$\mu = 17.75$

Generally the proportion that are not staying with parents is

$\r q = 1 - \r p$

= > $\r q = 0.858$

The standard deviation is mathematically evaluated as

$\sigma = \sqrt{n * \r p * \r q }$

$\sigma = \sqrt{ 125 * 0.142 * 0.858 }$

$\sigma = 3.90$

Given the n is large then we can use normal approximation to evaluate the probability as follows

$P(14 < X < 20 ) = P( \frac{ 14 - 17.75}{3.90}$

Now applying continuity correction

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < \frac{ X - \mu }{\sigma } < \frac{ 19.5 - 17.75}{3.90} )$

Generally

$\frac{ X - \mu }{\sigma } = Z ( The \ standardized \ value \ of X )$

$P(14 < X < 20 ) = P( \frac{ 13.5 - 17.75}{3.90} < Z< \frac{ 19.5 - 17.75}{3.90} )$

$P(14 < X < 20 ) = P( -1.0897 < Z< 0.449 } )$

$P(14 < X < 20 ) = P( Z< 0.449 ) - P(Z < -1.0897)$

So for the z - table

$P( Z< 0.449 ) = 0.67328$

$P(Z < -1.0897) = 0.13792$

$P(14 < X < 20 ) = 0.67328 - 0.13792$

$P(14 < X < 20 ) = 0.5354$

6 0

3 years ago

What is:<br> -6s²+-4.5=?

ddd [48]

Answer:

-6s² + - 4.5

= -6s² +(-4.5)

= -6s² - 4.5

3 0

2 years ago

Read 2 more answers

According to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation, there were over 71 million wildif

suter [353]

Answer:

The probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is P=0.412.

Step-by-step explanation:

We know the population proportion π=0.8, according to the 2011 National Survey of Fishing, Hunting, and Wildlife-Associated Recreation.

If we take a sample from this population, and assuming the proportion is correct, it is expected that the sample's proportion to be equal to the population's proportion.

The standard deviation of the sample is equal to:

$\sigma_s=\sqrt{\frac{\pi(1-\pi)}{N}}=\sqrt{\frac{0.8*0.2}{500}}=0.018$

With the mean and the standard deviaion of the sample, we can calculate the z-value for 0.79 and 0.81:

$z=\frac{p-\pi}{\sigma}=\frac{0.79-0.80}{0.018} =\frac{-0.01}{0.018} = -0.55\\\\z=\frac{p-\pi}{\sigma}=\frac{0.81-0.80}{0.018} =\frac{0.01}{0.018} = 0.55$

Then, the probability that between 79% and 81%of the 500 sampled wildlife watchers actively observed mammals in 2011 is:

$P(0.79$

3 0

3 years ago