Would appreciate some help here, thanks!

Jon finds a box of 350 postcards. He selects a random sample of 75 postcards and finds that 27 already have a stamp on them. Pre

DiKsa [7]

Answer:

126

Step-by-step explanation:

Let the total number of postcard with stamp on it =x

Out of 350 postcards, x has stamp on it.

Out of 75 postcards, 27 has stamp on it.

Expressing these as a ratio:

350:x=75:27

350/x=75/27

Cross multiply

75x=350*27

x=(350*27)/75

x= 126

Therefore, the the total number of postcards in the box with stamps on them is 126.

5 0

3 years ago

2a + 6 = 12<br> What does (A) equal?

Daniel [21]

Answer:

a = 3

Step-by-step explanation:

2a + 6 = 12

-6 -6

----------------

2a = 6

take six and divide by 2

a = 3

8 0

3 years ago

Read 2 more answers

The height of a tree at time t is given by a twice-differentiable function H, where H(t) is measured in meters and t is measured

il63 [147K]

A) Use the mean value theorem.

$H'(6)\approx\dfrac{H(7)-H(5)}{7-5}=\dfrac{11-6}2=\dfrac52\text{ meters/year}$

Since $H(t)$ gives the tree's height in meters at time $t$ , the value of $H'(6)$ informs us how quickly the tree is growing exactly after 6 years have passed. (i.e. the instantaneous rate of change of the tree's height)

b) We use the mean value theorem again. Observe that $H(5)-H(3)=6-2=4$ , and that $5-3=2$ . By the MVT, there must be some $3$ such that

$H'(t)=\dfrac42=2$

c) The average height of the tree is given by the integral

$\displaystyle\frac1{10-2}\int_2^{10}H(t)\,\mathrm dt$

If you can remember the formula for the area of a trapezoid, then this is pretty easy to compute. With five data points, you end up with four trapezoids constructed by the four adjacent subintervals. The "bases" are given by the values of $H(t)$ at each pair of endpoints, and the "heights" are the lengths of the subintervals. For the integral itself, we get

$\dfrac{2+1.5}2(3-2)+\dfrac{6+2}2(5-3)+\dfrac{11+6}2(7-5)+\dfrac{15+11}2(10-7)=\dfrac{263}4$

So the average height of the tree (in meters) is

$\displaystyle\frac1{10-2}\int_2^{10}H(t)\,\mathrm dt\approx\frac{263}{32}$

d) When $G=50$ , the diameter of the base can be determined to be

$50=\dfrac{100x}{1+x}\implies x=1$

We're told that $\dfrac{\mathrm dx}{\mathrm dt}=0.03$ . $G$ is a function of $x$ which is in turn a function of $t$ , so when we differentiate, we use the chain rule:

$\dfrac{\mathrm dG}{\mathrm dt}=\dfrac{\mathrm dG}{\mathrm dx}\cdot\dfrac{\mathrm dx}{\mathrm dt}$

$\implies\dfrac{\mathrm dG}{\mathrm dt}=\dfrac{100}{(1+x)^2}\cdot\dfrac3{100}=\dfrac3{(1+x)^2}$

When the height of the tree is 50 meters, we found the diameter to be 1 meter, so at this point

$\dfrac{\mathrm dG}{\mathrm dt}=\dfrac34\text{ meters/year}$

4 0

2 years ago

Read 2 more answers

the water works commission needs to know the mean household usage of water by the residents of a small town in gallons per day.

marin [14]

To determine the average water use, 109 people must be sampled.

1.9 gallons is the standard deviation.

E = 0.15 gallons maximum error

1.9 gallons on average

90% is the critical value.

1.645 is the 90% confidence interval.

sample size requirement,

n = (((z ÷ 2) × σ) ÷ E)²

n = (((1.645 ÷ 2) × 1.9) ÷ 0.15)²

n = ((0.8225 × 1.9) ÷ 0.15)²

n = (1.5627 ÷ 0.15)²

n = (10.418)²

n = 108.53 ≈ 109

As a result, the minimal sample size necessary to determine the mean water use = 109.

It is determined by dividing the average standard error by the squared of the sample size, and it decreases with increasing sample size. In other words, when the sample size is sufficiently big, the population mean approaches the population mean.

To learn more about the sample interval at

brainly.com/question/13664567?referrer=searchResults

#SPJ4

8 0

1 year ago

a storage tank has the dimensions shown below how many storage tanks are needed to contain 60000 cubic meters of liquid

Sladkaya [172]

First find the volume of that container

R = pi (r)^2h
R = pi (15.21)(12.4)
R = 592.5169408

60,000 / R = 101.2629275

102 storage tanks

4 0

3 years ago