B. f(x)=2x-4

C. f(x)=-3x+4

D f(x)= 3(2^x)-4

Answer:

**Answer:**

A. f(x)= 2(3^x)+4

**Step-by-step explanation:**

The linear equations of answer choices B and C will not have a horizontal asymptote. The exponential equation of choice D will have a horizontal asymptote at y=-4.

The appropriate choice is the exponential equation with 4 added:

f(x) = 2(3^x)+4

Measure the amgle of ced

N f(x) = 2x , x=5. What is the value of the function?

1/2(3/5x-8)+7/4(2/5x+4)

- It requires 1/4 of a credit to play a video game for one minute.a Emma has 7/8credits. Can she play for more or less than one minute? Explainhow you know

Answers in Scientific Notation.

N f(x) = 2x , x=5. What is the value of the function?

1/2(3/5x-8)+7/4(2/5x+4)

- It requires 1/4 of a credit to play a video game for one minute.a Emma has 7/8credits. Can she play for more or less than one minute? Explainhow you know

Answers in Scientific Notation.

1.5 I think would be the correct answer not sure

Juan is applying basic statistical principles in a chemistry laboratory by reviewing the standard deviation of the lab measurements and repeating his measurements multiple times to find a more accurate mean. The more Juan repeats his measurements, the closer he gets to a normal distribution or an accurate mean as per the central limit theorem.

In this chemistry laboratory scenario, you're dealing with a situation in **statistics** known as repeated measurements. Essentially, you are considering the standard deviation of the lab measurements, which is a typical measure of the dispersion of a set of values. The **standard deviation** is denoted by σ, and it is given as 10 milligrams.

When Juan repeats the measurement 4 times and records the mean of his measurements, he's using another common measure of central tendency, the arithmetic mean.

According to the central limit theorem in statistics, the distribution of the mean of a large number of independent, identically distributed variables will be approximately normal, regardless of the underlying distribution. In this case, as Juan repeats his measurements, the mean of these measurements is likely to be more accurate (closer to the true value) than a single measurement.

#SPJ6

The **standard **deviation a measure of dispersion in a data set, lower values indicating data points closer to the mean of the data set, and higher values indicating a wide range of the data points. The scenario discusses the calculation of standard deviation for repeated measurements, with the standard error calculated as the original standard deviation divided by the square root of the number of measurements.

The subject matter of the question pertains to statistical concepts, primarily the standard deviation. In statistics, the standard deviation is a measure of the amount of variation or dispersion in a data set. A low standard deviation indicates that the data points tend to be close to the mean of the data set, while a high standard deviation indicates that the data points are spread out over a wider range.

In the scenario provided, Juan makes a measurement in a chemistry lab and the standard **deviation **of the students' lab measurements is 10mg. He repeats the measurement 4 times and records the mean of his 4 measurements. When you repeat a measurement multiple times and take the mean, the standard deviation of the mean tends to be smaller than the standard deviation of the individual measurements. In statistical terms, the standard deviation of the mean, also known as the standard error, is given by the original standard deviation σ divided by the square root of the number of measurements n. In this case, n is 4, so the standard error would be σ/√n = 10mg/√4 = 5mg.

#SPJ11

**Answer:**

(-9, 1)

**Step-by-step explanation:**

translating UP is to increase on the y axis, so -4+5=1

translating to the LEFT is to subtract on the x axis so-2-7=-9

-9, 1 I think is the correct answer

(a)The propability that event will occur is....(TYPE AN INTEGER OR DECIMAL ROUNDED TO THE NEAREST THOUSANDTH AS NEEDED.)

(b)The propability thet the event will not occur is...(TYPE AN INTEGER OR DECIMAL ROUNDED TO THE NEAREST THOUSANDTH AS NEEDED)

**Answer:**

A) The probability that the event will occur

B)The probability that the event will not occur =

**Step-by-step explanation:**

We are given that The odds of event occurring are 1:6.

So, Number of successful events = 1

Number of unsuccessful events = 6

So, Total events = 6+1=7

a)the probability that the event will occur=

The probability that the event will occur

b)The probability that the event will not occur =

The probability that the event will not occur =

Answer:

q = 0.105uC

Step-by-step explanation:

We can determine the force on one ball by assuming two balls are stationary, finding the E field at the lower right vertex and calculate q from that.

Considering the horizontal and vertical components.

First find the directions of the fields at the lower right vertex. From the lower left vertex the field will be at 0° and from the top vertex, the field will be at -60° or 300° because + charge fields point radially outward in all directions. The distances from both charges are the same since this is an equilateral triangle. The fields have the same magnitude:

E=kq/r²

Where r = 20cm

= 20/100

= 0.2m

K = 9.0×10^9

9.0×10^9 × q /0.2²

9.0×10^9/0.04

2.25×10^11 q

These are vector fields of course

Sum the horizontal components

Ecos0 + Ecos300 = E+0.5E

= 1.5E

Sum the vertical components

Esin0 + Esin300 = -E√3/2

Resultant = √3E at -30° or 330°

So the force on q at the lower right corner is q√3×E

The balls have two forces, horizontal = √3×E×q

and vertical = mg, therefore if θ is the angle the string makes with the vertical tanθ = q√3E/mg

mg×tanθ = q√3E.

..1

Then θ will be...

Since the hypotenuse = 80cm

80cm/100

= 0.8m

The distance from the centroid to the lower right vertex is 0.1/cos30 =

0.1/0.866

= 0.1155m

Hence,

0.8×sinθ = 0.1155

Sinθ = 0.1155/0.8

Sin θ = 0.144375

θ = arch sin 0.144375

θ = 8.3°

From equation 1

mg×tanθ = q√3E

g = 9.8m/s^2

m = 3.0g = 0.003kg

0.003×9.8×tan(8.3)

0.00428 = q√3E

0.00428 = q×1.7320×E

Where E=kq/r²

Where r = 0.2m

0.0428 = kq^2/r² × 1.7320

K = 9.0×10^9

0.0428/1.7320 = 9.0×10^9 × q² / 0.2²

0.02471×0.04 = 9.0×10^9 × q²

0.0009884 = 9.0×10^9 × q²

0.0009884/9.0×10^9 = q²

q² = 109822.223

q = √109822.223

q = 0.105uC

Hope this helps, let me know if it doesn't make sense