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2 years ago

Help, 100 pts + brainlist for first person to answer

Mathematics

2 answers:

Blababa [14]2 years ago

8 0

Answer:

About 4.

Step-by-step explanation:

It is actaully 3.95 but if you were rounding and in fraction form, it would be four! Hope this helps!

aalyn [17]2 years ago

6 0

Answer:

3.95151515…………………………………………

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Find the equation of a line passing through (3, -5) that is parallel to 2x + 6y =10

PtichkaEL [24]

Answer:

The equation of the line would be y = -1/3x - 4

Step-by-step explanation:

In order to find this, we first need to find the slope of the original line. We do this by solving for y.

2x + 6y = 10

6y = -2x + 10

y = -1/3x + 5/3

Now that we see the slope as -1/3, we know the new line will have the same slope thanks to the definition of parallel lines. So, we can use this slope and the point in point-slope form to find the equation.

y - y1 = m(x - x1)

y - -5 = -1/3(x - 3)

y + 5 = -1/3x + 1

y = -1/3x - 4

7 0

3 years ago

Thirty-seven percent of all Americans drink bottled water more than once a week (Natural resources Defense Council, December 4,

lesantik [10]

Answer:

a) The sampling distribution for n=540 has a mean sample proportion of p=0.37 and a standard deviation of σs=0.0208.

b) probability = 0.99998

d) probability = 0.99874

e) You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

Step-by-step explanation:

a) We have a known population proportion π=0.37 and we have to describe the sampling distribution when the sample size is n=540.

The mean sample proportion is expected to be the same as the population proportion:

$\bar p = \pi = 0.37$

The standard deviation of the sampling will be the population standard deviation divided by the square root of the sample size:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{540}}=\sqrt{0.000431667}=0.0208$

Then, we can say that the sampling distribution will have a p=0.37 and a standard deviation σs=0.0208.

b) We have to calculate the probability that the sample proportion will be within 0.09 of the population proportion.

We can calculate the z-value as:

$z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0208}=4.3269\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0208}=-4.3269$

As the distribution is symmetrical, we can calculate the probabilty that he sample proportion will be within 0.09 of the population proportion as:

$P(|p-\bar p|$

probability = 0.99998

d. Now the sample is smaller (n=300), so the standard deviation of the samping distribution:

$\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{300}}=\sqrt{0.000777}=0.0279$

We have to recalculate the z-scores:

$z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0279}=3.2258\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0279}=-3.2258$

And the probability is:

$P(|p-\bar p|$

probability = 0.99874

e. The increase in sample size is 80%

$\Delta n\%=\dfrac{n_1}{n_2}-1=\dfrac{540}{300}-1=1.8-1=0.8=80\%$

and the increase in probability is 0.12%

$\Delta P\%=\dfrac{P_1}{P_2}-1=\dfrac{0.99998}{0.99874}-1=1.0012-1=0.0012=0.12\%$

You gain 0.12% in probability for an increase of 80% in sample size.

The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).

7 0

3 years ago

Find the integral of √(x² +4) W.R.T x

Allushta [10]

Answer:

$\frac{x}{2} *\sqrt{x^{2} +4}$ + $\frac{1}{2}$ *LN(| $\frac{x+\sqrt{x^{2} +4} }{2}$ |) +C

Step-by-step explanation:

we will have to do a trig sub for this

use x=a*tanθ for sqrt(x^2 +a^2) where a=2

x=2tanθ, dx= 2 sec^2 (θ) dθ

this turns $\int\limits {\sqrt{x^{2}+4 } } \, dx$ into integral(sqrt( [2tanθ]^2 +4) * 2sec^2 (θ) )dθ

the sqrt( [2tanθ]^2 +4) will condense into 2sec^2 (θ) after converting tan^2(θ) into sec^2(θ) -1

then it simplifies into integral(4*sec^3 (θ)) dθ

you will need to do integration by parts to work out the integral of sec^3(θ) but it will turn into (1/2)sec(θ)tan(θ) + (1/2) LN(|sec(θ)+tan(θ)|) +C

then you will need to rework your functions of θ back into functions of x

tanθ will resolve back into $\frac{x}{2}$ (see substitutions) while secθ will resolve into $\frac{\sqrt{x^{2} +4} }{2}$

sec(θ)= $\frac{\sqrt{x^{2} +4} }{2}$ is from its ratio identity of hyp/adj where the hyp. is $\sqrt{x^{2} +4}$ and adj is 2 (see tan(θ) ratio)

after resolving back into functions of x, substitute ratios for trig functions:

= $\frac{x}{2} *\sqrt{x^{2} +4}$ + $\frac{1}{2}$ *LN(| $\frac{x+\sqrt{x^{2} +4} }{2}$ |) +C

3 0

3 years ago

Repost because someone answered without actually giving an answer. What is the mode of the data shown in the table? Scores: 5 12

nordsb [41]

Answer:

Option (D) : 13

Step-by-step explanation:

Mode is the observation which occurred the most number of time.

Therefore, mode = 13

8 0

3 years ago

Gina measured the amount of water her friends drink each day in gallons. The line plot shows the measurements.

vova2212 [387]

Line plots are used to represent data using lines and dots

The difference between the greatest amount of water and the least is 1/2

From the line plot (see attachment), we have the following parameters:

$Least = 8\frac 28$

$Greatest = 8\frac 68$

<h3>Calculating the difference</h3>

The difference (d) is then calculated as:

$d =Greatest -Least$

So, we have:

$d =8\frac 68 -8\frac 28$

Subtract the common terms (8)

$d =\frac 68 -\frac 28$

Take LCM

$d =\frac {6-2}8$

$d =\frac {4}8$

Reduce the fraction

$d =\frac {1}2$

Hence, the difference between the greatest amount of water and the least is 1/2