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3 years ago

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Mathematics

1 answer:

likoan [24]3 years ago

3 0

Answer:

Step-by-step explanation:

https://tex.z-dn.net/?f=24x%5E%7B5%7Dy%5E%7B4%7D%5Csqrt%7B7x%5E%7B2%7Dy%7D

this is correct

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In 2012 there were approximately 8,950 public libraries in the United States. A survey found that 76% of those libraries offered

choli [55]

Answer:

6,440

Step-by-step explanation:

i just did the math for this

5 0

3 years ago

Parallel perpendicular or neither

Elenna [48]

Answer:

parallel

Step-by-step explanation:

4-2/-9-0= 2/-9

6-8/4+5= -2/9

4 0

3 years ago

A tank contains 250 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu

VashaNatasha [74]

Answer:

$A(t)=250-210e^{-\frac{t}{50}}$

Step-by-step explanation:

We are given that

Volume,V=250 L

Mass of salt=m=40 g

Brine containing 1 g per liter

Pumped into the tank at the rate=5 L/min

We have to find the number A(t) of grams of salt in the tank at time t.

Rate of change of salt in the tank

$\frac{dA}{dt}=$ Rate in-Rate out

Rate in=5 L/min

Rate out= $\frac{A}{250}\times 5=\frac{A}{50}$ L/min

$\frac{dA}{dt}=5-\frac{A}{50}=\frac{250-A}{50}$

$\int \frac{50dA}{250-A}=\int dt$

$-50ln(250-A)=t+C$

Using the formula

$\int \frac{dx}{x}=ln x$

A=40 and t=0

$-50ln(250-40)=0+C$

$-50ln(210)=C$

Substitute the value

$-50ln(250-A)=t-50ln(210)$

$-50ln(250-A)+50ln(210)=t$

$50ln\frac{210}{250-A}=t$

$t=50ln\frac{210}{250-A}$

$\frac{t}{50}=ln\frac{210}{250-A}$

$\frac{210}{250-A}=e^{\frac{t}{50}$

$\frac{250-A}{210}=e^{-\frac{t}{50}}$

$250-A=210e^{-\frac{t}{50}}$

$A(t)=250-210e^{-\frac{t}{50}}$

8 0

4 years ago

A researcher surveyed 150 high school students and found that 68% played a musical insturment. What would be a reasonable range

maria [59]

Answer:

The reasonable range for the population mean is (61%, 75%).

Step-by-step explanation:

The interval estimate of a population parameter is an interval of values that consist of the values within which the true value of the parameter lies with a certain probability.

The mean of the sampling distribution of sample proportion is, $\hat p$ .

One of the best interval estimate of population proportion is the 95% confidence interval for proportion,

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

Given:

n = 150

$\hat p$ = 0.68

The critical value of <em>z</em> for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.68\pm1.96\sqrt{\frac{0.68(1-0.68)}{150}}\\\\=0.68\pm 0.0747\\\\=(0.6053, 0.7547)\\\\\approx (0.61, 0.75)$

Thus, the reasonable range for the population mean is (61%, 75%).

5 0

4 years ago

Determine the graph of the polar equation r = 25/10-25 sin(theta).

ludmilkaskok [199]

Answer:

graph A

Step-by-step explanation:

A graphing calculator can plot this function for you.

If you divide numerator and denominator by 25, you get ...

r = 1/(0.4 -sin(θ))

This will have asymptotes at θ = arcsin(0.4). At those values of θ, the value of r will change sign, so the graph cannot be a parabola or ellipse.

The graph of the hyperbola is the appropriate choice.

7 0

3 years ago