A cell site is a site where electronic communications equipment is placed in a cellular network for the use of mobile phones. The numbers y of cell sites from 1985 through 2014 can be modeled by
y = 340,110/
1 + 377e−0.259t
where t represents the year, with
t = 5 corresponding to 1985.
Use the model to find the numbers of cell sites in the years 1998, 2003, and 2006

Answers

Answer 1

Answer:

(a) 74553

(b) 172120

(c) 234802

Step-by-step explanation:

Given

[tex]y = \frac{340110}{1 + 377e^{-0.259t}}[/tex]

Solving (a): 1998

Year 1998 means that:

[tex]t =1998 - 1980[/tex]

[tex]t =18[/tex]

So, we have:

[tex]y = \frac{340110}{1 + 377e^{-0.259*18}}[/tex]

[tex]y = \frac{340110}{1 + 377e^{-4.662}}[/tex]

[tex]y = \frac{340110}{1 + 3.562}[/tex]

[tex]y = \frac{340110}{4.562}[/tex]

[tex]y = 74553[/tex] --- approximated

Solving (b): 2003

Year 2003 means that:

[tex]t = 2003 - 1980[/tex]

[tex]t =23[/tex]

So, we have:

[tex]y = \frac{340110}{1 + 377e^{-0.259*23}}[/tex]

[tex]y = \frac{340110}{1 + 377e^{-5.957}}[/tex]

[tex]y = \frac{340110}{1 + 0.976}[/tex]

[tex]y = \frac{340110}{1.976}[/tex]

[tex]y = 172120[/tex] --- approximated

Solving (c): 2006

Year 2006 means that:

[tex]t = 2006 - 1980[/tex]

[tex]t =26[/tex]

So, we have:

[tex]y = \frac{340110}{1 + 377e^{-0.259*26}}[/tex]

[tex]y = \frac{340110}{1 + 377e^{-6.734}}[/tex]

[tex]y = \frac{340110}{1 + 0.4485}[/tex]

[tex]y = \frac{340110}{1.4485}[/tex]

[tex]y = 234802[/tex] --- approximated


Related Questions

Can you help me answer this question? Screenshot is added.

Answers

9514 1404 393

Answer:

  (c)

Step-by-step explanation:

  [tex]\displaystyle\sqrt[3]{xy^5}\sqrt[3]{x^7y^{17}}=\sqrt[3]{x^{1+7}y^{5+17}}=\sqrt[3]{x^6x^2y^{21}y}=\sqrt[3]{x^6y^{21}}\sqrt[3]{x^2y}\\\\=\boxed{x^2y^7\sqrt[3]{x^2y}}[/tex]

What is graph for the equation y=-4x+1

Answers

Answer: The line starts at 1 positive, then from there go -4 (so go to the left) then 1 down from that point.

Step-by-step explanation: the problem is supposed to have been Y= -4/1 +1

Order the following decimals. State your method of choice and your reasons for choosing it. Explain how you know this order is accurate.

Answers

Answer:

.40 is the greatest .350 is the second greatest and last but not least .3456 is the lowest

Step-by-step explanation:

The population of a bacteria colony is growing exponentially, doubling every 6 hours. If there are 150 bacteria currently present, how many (to the nearest ten bacteria) will be present in 10 hours

Answers

Answer:

If rounded to the nearest 10 bacteria, then it would be 500 bacteria.

Step-by-step explanation:

First multiply 150 by two in order to get 300, that leaves 4 hours to figure out. From there you can figure out the rest by seeing that 4 is 2/3 of 6. I converted it into the decimal number .66. Multiply 300 by .66 to get 198 and then add it to 300 to get 498. Then just round it up to the nearest 10 bacteria which leaves you with the final answer of 500 bacteria.

The length of the box is 15 centimeters, the breadth of the box is 20 centimeter, the height of a box, 20 centimeter fine its volume. Step by step​

Answers

Answer:

volume=length×width×height

v=15×20×20

v=6000

3x+4 number of terms

Answers

9514 1404 393

Answer:

  2

Step-by-step explanation:

In this expression, the terms are the parts of the sum. They are 3x and 4. There are 2 terms.

what graph shows the solution to the equation below log3(x+2)=1

Answers

Answer:

The solution to the equation  log3(x+2)=1 is given by x=1

Step-by-step explanation:

We are given that

[tex]log_3(x+2)=1[/tex]

We have to find the graph which shows the  solution to the equation log3(x+2)=1.

[tex]log_3(x+2)=1[/tex]

[tex]x+2=3^1[/tex]

Using the formula

[tex]lnx=y\implies x=e^y[/tex]

[tex]x+2=3[/tex]

[tex]x=3-2[/tex]

[tex]x=1[/tex]

Type the correct answer in each box. Use numerals instead of words.
Multiply the expressions.



If a = 1, find the values of b, c, and d that make the given expression equivalent to the expression below.

Answers

Answer:

a=1, b=9, c=-2, d=4

Step-by-step explanation:

What’s the equation of the line

Answers

Answer:

[tex]y = - \frac{1}{3}x + 5[/tex]

Step-by-step explanation:

Consider two points through which the line passes.

Let it be ( 0 , 5 ) and ( 6 , 3 )

Step 1 : Find slope

    [tex]Slope, m = \frac{y_2 - y_ 1 }{x_2 - x_1}[/tex]

                 [tex]= \frac{3-5}{6-0} \\\\=\frac{-2}{6}\\\\= -\frac{1}{3}[/tex]

Step 2 : Find the equation of the line passing through the points.

       [tex]( y - y_1) = m (x - x_1)\\\\(y - 5) = -\frac{1}{3} ( x - 0) \\\\y = -\frac{1}{3}x + 5[/tex]

Order these numbers from least to greatest.
5.772 , 11/2, 5 6/11, 5.77

Answers

Answer:

6/11, 11/2, 5.77, 5.772

Step-by-step explanation:

If y- 1 equals 10 then y

Answers

Answer:

11

Step-by-step explanation:

y-1=10

Any figure that crosses equal sign, the operational sign changes.

y=10+1

y= 11

A closed, rectangular-faced box with a square base is to be constructed using only 36 m2 of material. What should the height h and base length b of the box be so as to maximize its volume

Answers

Answer:

[tex]b=h=\sqrt{6}[/tex] m

Step-by-step explanation:

Let

Bas length of box=b

Height of box=h

Material used in constructing of box=36 square m

We have to find the height h and base length b of the box to maximize the volume of box.

Surface area of box=[tex]2b^2+4bh[/tex]

[tex]2b^2+4bh=36[/tex]

[tex]b^2+2bh=18[/tex]

[tex]2bh=18-b^2[/tex]

[tex]h=\frac{18-b^2}{2b}[/tex]

Volume of box, V=[tex]b^2h[/tex]

Substitute the values

[tex]V=b^2\times \frac{18-b^2}{2b}[/tex]

[tex]V=\frac{1}{2}(18b-b^3)[/tex]

Differentiate w. r.t b

[tex]\frac{dV}{db}=\frac{1}{2}(18-3b^2)[/tex]

[tex]\frac{dV}{db}=0[/tex]

[tex]\frac{1}{2}(18-3b^2)=0[/tex]

[tex]\implies 18-3b^2=0[/tex]

[tex]\implies 3b^2=18[/tex]

[tex]b^2=6[/tex]

[tex]b=\pm \sqrt{6}[/tex]

[tex]b=\sqrt{6}[/tex]

The negative value of b is not possible because length cannot be negative.

Again differentiate w.r.t b

[tex]\frac{d^2V}{db^2}=-3b[/tex]

At  [tex]b=\sqrt{6}[/tex]

[tex]\frac{d^2V}{db^2}=-3\sqrt{6}<0[/tex]

Hence, the volume of box is maximum at [tex]b=\sqrt{6}[/tex].

[tex]h=\frac{18-(\sqrt{6})^2}{2\sqrt{6}}[/tex]

[tex]h=\frac{18-6}{2\sqrt{6}}[/tex]

[tex]h=\frac{12}{2\sqrt{6}}[/tex]

[tex]h=\sqrt{6}[/tex]

[tex]b=h=\sqrt{6}[/tex] m

what is the sum of a 7 term geometric series if the first term is 6 the last term is -24576 and the common ratio is -4

Answers

Answer:

Sum = 19,662

Step-by-step explanation:

Given that this is a finite geometric series (meaning it stops at a specific term or in this case -24,576), we can use this formula:  

[tex]\frac{a(1-r^n)}{1-r}[/tex], where a is the first term, r is the common ratio, and n is the number of terms.

Substituting for everything and simplifying gives us:

[tex]\frac{6(1-(-4)^7)}{1-(-4)} \\\\\frac{6(16385}{5}\\ \\\frac{98310}{5}\\ \\19662[/tex]

a student takes two subjects A and B. Know that the probability of passing subjects A and B is 0.8 and 0.7 respectively. If you have passed subject A, the probability of passing subject B is 0.8. Find the probability that the student passes both subjects? Find the probability that the student passes at least one of the two subjects

Answers

Answer:

0.64 = 64% probability that the student passes both subjects.

0.86 = 86% probability that the student passes at least one of the two subjects

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

In which

P(B|A) is the probability of event B happening, given that A happened.

[tex]P(A \cap B)[/tex] is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Passing subject A

Event B: Passing subject B

The probability of passing subject A is 0.8.

This means that [tex]P(A) = 0.8[/tex]

If you have passed subject A, the probability of passing subject B is 0.8.

This means that [tex]P(B|A) = 0.8[/tex]

Find the probability that the student passes both subjects?

This is [tex]P(A \cap B)[/tex]. So

[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]

[tex]P(A \cap B) = P(B|A)P(A) = 0.8*0.8 = 0.64[/tex]

0.64 = 64% probability that the student passes both subjects.

Find the probability that the student passes at least one of the two subjects

This is:

[tex]p = P(A) + P(B) - P(A \cap B)[/tex]

Considering [tex]P(B) = 0.7[/tex], we have that:

[tex]p = P(A) + P(B) - P(A \cap B) = 0.8 + 0.7 - 0.64 = 0.86[/tex]

0.86 = 86% probability that the student passes at least one of the two subjects

The weights of certain machine components are normally distributed with a mean of 5.19 ounces and a standard deviation of 0.05 ounces. Find the two weights that separate the top 8% and the bottom 8%. These weights could serve as limits used to identify which components should be rejected

Answers

Answer:

The  weight that separate the top 8% by 5.2605 and the weight that separate bottom 8% by 5.1195.

Step-by-step explanation:

We are given that

Mean,[tex]\mu=5.19[/tex]

Standard deviation,[tex]\sigma=0.05[/tex]

We have to find the two weights that separate the top 8% and the bottom 8%.

Let x1 and x2 the two weights that separate the top 8% and the bottom 8%.

Z-value for p-value =0.08 =[tex]-1.41[/tex]

For 8% bottom

[tex]Z=\frac{x_1-\mu}{\sigma}=-1.41[/tex]

[tex]\frac{x_1-5.19}{0.05}=-1.41[/tex]

[tex]x_1-5.19=-1.41\times 0.05[/tex]

[tex]x_1=-1.41\times 0.05+5.19[/tex]

[tex]x_1=5.1195[/tex]

For 8% top

p-Value=1-0.08=0.92

Z- value=1.41

Now,

[tex]\frac{x_2-5.19}{0.05}=1.41[/tex]

[tex]x_2-5.19=1.41\times 0.05[/tex]

[tex]x_2=1.41\times 0.05+5.19[/tex]

[tex]x_2=5.2605[/tex]

(x1,x2)=(5.1195,5.2605)

Draw a line representing the "rise" and a line representing the "run" of the line. State the slope of the line in simplest form.

Answers

Answer:

See attachment showing the rise and run

Slope = 1

Step-by-step explanation:

In the diagram attached below, the rise is represented by the blue line, while the run is represented by the red line.

Rise = 4 units

Run = 4 units

It's a positive slope because the line slopes upwards from left to right

Slope = rise/run = 4/4

Slope = 1

Given sets X, Y, Z, and U, find the set Xn(X - Y) using the listing method.
X = {d, c, f, a}
Y = {d, e, c}
Z ={e, c, b, f, g}
U = {a, b, c, d, e, f, g}

Answers

Answer:

{f, a}

Step-by-step explanation:

Given the sets:

X = {d, c, f, a}

Y = {d, e, c}

Z ={e, c, b, f, g}

U = {a, b, c, d, e, f, g}

To obtain the set X n (X - Y)

We first obtain :

(X - Y) :

The elements in X that are not in Y

(X - Y) = {f, a}

X n (X - Y) :

X = {d, c, f, a} intersection

(X - Y) = {f, a}

X n (X - Y) = elements in X and (X - Y)

X n (X - Y) = {f, a}

Use the given graph of f to state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
The x y-coordinate plane is given.
The function enters the window in the second quadrant, goes up and right becoming less steep, crosses the y-axis at approximately y = 3.2, changes direction at the approximate point (0.7, 3.3), goes down and right becoming more steep, and stops at the closed point (2, 3).
The function starts again at the open point (2, 1), goes up and right becoming more steep, goes up and right becoming less steep, passes through the open point (4, 4), changes direction at the approximate point (4.2, 4.1), goes down and right becoming more steep, and exits the window in the first quadrant.
(a) lim x → 2− f(x)
(b) lim x → 2+ f(x)
(c) lim x → 2 f(x)
(d) f(2)
(e) lim x → 4 f(x)(f) f(4)

Answers

Answer:

Hence the answer is given as follows,

Step-by-step explanation:

Graph of y = f(x) given,

(a) [tex]\lim_{x\rightarrow 2^{-}}f(x)=3[/tex]

(b) [tex]\lim_{x\rightarrow 2^{+}}f(x)=1[/tex]

(c) [tex]\lim_{x\rightarrow 2}f(x)= DNE \left \{ \therefore \lim_{x\rightarrow 2^{-}} f(x)\neq \lim_{x\rightarrow 2^{+}}f(x) \right.[/tex]

(d) [tex]f(2)=3[/tex]

(e) [tex]\lim_{x\rightarrow 4}f(x) = 4[/tex]

(f) [tex]f(4)= DNE.[/tex]{ Hole in graph}

Hence solved.

ASAP What is the rule for this relation? i will give brainliest

Answers

Answer:

your selected answer is right

A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are given below. At the 1% level of significance, test the claim that the sensory measurements are lower after hypnotism (scores are in cm. on a pain scale). Assume sensory measurements are normally distributed. Note: You do not need to type these values into Minitab Express; the data file has been created for you.Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0

Answers

Answer:

sensory measurement are lower after hypnotism

Step-by-step explanation:

Given the data :

Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6

After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0

The difference ;

After - Before, d = 0.2, - 4.1, - 1.6, - 1.8, - 3.2, - 2, - 2.9, - 9.6

Hypothesis :

H0 : μd = 0

H0 : μ < 0

The test statistic ;

T = μd / sd/√n

Where, xd = mean of difference

sd = standard deviation of difference

n = sample size

Mean of difference, μd = Σx/n = - 3.13

Standard deviation of difference, sd = 2.91

T = - 3.13 / 2.91/√8

T = - 3.13 / 1.0288403

T = - 3.042

α = 0.01

The Pvalue using a Pvalue calculator ;

Degree of freedom, df = n - 1 ; 8-1 = 7

Pvalue(-3.042, 7) = 0.00939

Pvalue < α ; we reject the null and conclude that sensory measurement are lower after hypnotism

what is the value of the expression 5²5 ​

Answers

Answer:

5^2×5

=25×5

=125

hope this will help you

Answer:

125

Step-by-step explanation:

hope this helps you

=25×5

=125

The awnser for this question

Answers

the answer is 52

c=cost, which is given to be $20

plug that in to the given equation, c=12.70+0.14t

so it’ll be
20 = 12.70 + 0.14t

subtract 12.70 from each side

7.30 = 0.14t

divide both sides by 0.14 to get t by itself

52.1 = t

round that to 52

The accompanying data represent the homework scores for material for a random sample of students in a college algebra course.
36
47
54
58
60
66
66
68
69
70
72
75
77
77
78
78
78
79
79
79
79
79
80
82
84
85
86
86
86
87
89
89
91
92
92
93
93
94
96
99
​(a) Construct a relative frequency distribution with a lower class limit of the first class equal to 30 and a class width of 10.
(b) What is the probability a randomly selected student fails the homework​ (scores less than​ 70)? (The standard deviation is 13.64)
Simplify your answer to two decimal​ places.

Answers

Answer:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

[tex]P(x < 70) = 0.225[/tex]

Step-by-step explanation:

Given

[tex]Lower = 30[/tex]

[tex]Width = 10[/tex]

Solving (a): The relative frequency table

First, we construct the frequency table using the given parameters.

[tex]\begin{array}{cc}{Class} & {Frequency} &{30-39} & {1} & {40-49} & {1} & {50 - 59} & {2} & {60 - 69} & {5} & {70 - 79} & {13} & {80 - 89} & {10} & {90 - 99} & {8} & {Total} & {40}\ \end{array}[/tex]

The relative frequency (RF) is calculated as:

[tex]RF = \frac{Frequency}{Total}[/tex]

Using the above formula to calculate the relative frequency, the relative frequency table is:

[tex]\begin{array}{ccc}{Class} & {Frequency} & {Relative\ Frequency} &{30-39} & {1} & {0.025} & {40-49} & {1} & {0.025} & {50 - 59} & {2} & {0.050} & {60 - 69} & {5} & {0.125} & {70 - 79} & {13} & {0.325} & {80 - 89} & {10} & {0.250} & {90 - 99} & {8} & {0.200} &{Total} & {40} & {1}\ \end{array}[/tex]

Solving (b): [tex]P(x < 70)[/tex]

To do this, we add up the relative frequencies of classes less than 70.

i.e.

[tex]P(x < 70) = [30 - 39] + [40 - 49] + [50 - 59] + [60 - 69][/tex]

So, we have:

[tex]P(x < 70) = 0.025 + 0.025 + 0.050 + 0.125[/tex]

[tex]P(x < 70) = 0.225[/tex]

Please help me quick I’ll give brainliest

Answers

I think c!!!!!!!!!!!!!!!!!!!!
the answer is D
because the first number positive goes left and the next one goes up because its positive ( hard to explain but(

The number of measles cases increased 26.3% to 321 cases this year. What was the number of cases prior to the increase? Express your answer rounded correctly to the nearest whole number.

Answers

Answer:

The right answer is "[tex]x\simeq 254[/tex]".

Step-by-step explanation:

Let the number of earlier case will be "x".

Now,

⇒ [tex]x+x\times \frac{26.3}{100}=321[/tex]

or,

⇒ [tex]x+x\times 0.263=321[/tex]

By taking "x" common, we get

⇒   [tex]x(1+0.263)=321[/tex]

⇒                    [tex]x=\frac{321}{1.263}[/tex]

⇒                       [tex]=254.15[/tex]

or,

⇒                    [tex]x\simeq 254[/tex]

if triangle TAN has vertices T(0, 2), A(-1,3), and N(-2,-4), which of the following coordinates is N' of the dilation from the origin using the scale factor 3?

Answers

Answer:

(-6,-12)

Step-by-step explanation:

A dilation makes a figure gets bigger so just multiply 3 to point N to find N prime.

[tex] - 2 \times 3 = - 6[/tex]

[tex] - 4 \times 3 = - 12[/tex]

So our new coordinates is

(-6,-12)

Answer:

(-6,-12)

Step-by-step explanation:

A dilation makes a figure gets bigger so just multiply 3 to point N to find N prime.

So our new coordinates is

(-6,-12)

Step-by-step explanation:

15×115-(-3)}(4-4)÷3{5+(-3)×(-6​

Answers

Answer:

15×115+3{0÷3}5-3×(-6)

15×115+3of 0 of 5-3×(-6)

15×115+0 of 5-3×(-6)

15×115+0+18

1725+0+18

1743

A display case of toy rings are marked 5 for $1. If Zach wants to buy 50 toy rings, how much will Zach spend (not including tax)

Answers

Answer:

$10

Step-by-step explanation:

5 toys = $1

Zach wants 50 of these

50 ÷ 5 = 10

10 x 1 = 10

= $10

Answer:

10 dollars

Step-by-step explanation:

We can use a ratio to solve

5 rings     50 rings

----------  = --------------

1 dollar         x dollars

Using cross products

5*x = 1 * 50

5x = 50

Divide by 5

5x/5 = 50/5

x = 10

Find the area of the shaded region in terms of .


Please help :)

Answers

9514 1404 393

Answer:

  50π cm²

Step-by-step explanation:

The radius of the larger circle is 10 cm, so its area is ...

  A = πr² = π(10 cm)² = 100π cm²

The area of each smaller circle is ...

  A = π(5 cm)² = 25π cm²

Then the shaded area is ...

  shaded = large circle - 2 × small circle

  shaded = 100π cm² - 2(25π cm²) = 50π cm²

4b^2+300=0 this is a quadratic equation that I am trying to solve including any solutions with imaginary numbers I will include a picture ​

Answers

Answer:

b= 5i square root of 3

b = -5i square root of 3

Step-by-step explanation:

4b^2+300=0

4b^2 = -300

b^2 = -75

b = square root of -75

b = -75^1/2

^1/2 means square root

b = 25^1/2 * 3^1/2 * i

b= 5i square root of 3

b = -5i square root of 3

Other Questions
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