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arlik [135]
4 years ago
8

What is the value of x3 + 4, when x = 6?

Mathematics
2 answers:
storchak [24]4 years ago
6 0
Well let’s remember the rules of pemdas.

Parentheses
Exponents
Multiplication
Division
Addition
Subtraction

Since 6 has a exponent of 3 we could do 6^3 before adding 4. 6^3= 216

So then we would have 216+4.

That would equal 220. The value when x=6 is 220.
tresset_1 [31]4 years ago
5 0
Plug in 6 for x
3(6) + 4 = 22
The value if 22, or you can write it like:
f(6) = 22
You might be interested in
The length of each side of a square was decreased by 2 inches so the perimeter is now 49 inches. What was the original length of
aleksley [76]

The Original length of each side of square was 14.25 inches.

Step-by-step explanation:

Perimeter of a square is given by:

Perimeter = 4 * side

Let x be the original length of each side of square

So

decreasing 2 inch will mean

x-2

Perimeter of square after reducing 2 inches from each side

4(x-2) = 49\\4x-8 = 49\\4x = 49+8\\4x = 57\\\frac{4x}{4} = \frac{57}{4}\\x = 14.25

So,

The Original length of each side of square was 14.25 inches.

Keywords: Perimeter, square

Learn more about perimeter at:

  • brainly.com/question/5922969
  • brainly.com/question/6013189

#LearnwithBrainly

5 0
3 years ago
The rectangle below has an area of x2 – 16 square<br> meters and a width of x + 4 meters.
Anna71 [15]

Answer:

What is the question? do you need the area? :)

Step-by-step explanation:

8 0
3 years ago
Find the circumference of the circle with the given radius or diameter. (Use 3.14 for π.) Diameter is 10cm
AlladinOne [14]

Answer:


Step-by-step explanationtake the diameter, 10cm and multiply by pi (3.14) and the answer is 31.4

5 0
3 years ago
A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard dev
Nesterboy [21]

Answer:

There is a 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 201.9-cm and a standard deviation of 2.1-cm. This means that \mu = 201.9, \sigma = 2.1.

For shipment, 9 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

By the Central Limit Theorem, since we are using the mean of the sample, we have to use the standard deviation of the sample in the Z formula. That is:

s = \frac{\sigma}{\sqrt{n}} = \frac{2.1}{\sqrt{9}} = 0.7

This probability is 1 subtracted by the pvalue of Z when X = 204.1.

Z = \frac{X - \mu}{\sigma}

Z = \frac{204.1 - 201.9}{0.7}

Z = 3.14

Z = 3.14 has a pvalue of 0.9992. This means that there is a 1-0.9992 = 0.0008 = 0.08% probability that the average length of a randomly selected bundle of steel rods is greater than 204.1-cm.

3 0
3 years ago
Suppose flights on standard routes between two given cities use on average 3,087 gallons of kerosene with standard deviation of
Radda [10]

Answer:

15.866%

Step-by-step explanation:

We solve using z score formula

z = (x-μ)/σ, where

x is the raw score = 3,282 gallons of kerosene

μ is the population mean = 3,087 gallons of kerosene

σ is the population standard deviation = 195 gallons

For x > 3282 gallons

z = 3282 - 3087/195

z = 1

Probability value from Z-Table:

P(x<3282) = 0.84134

P(x>3282) = 1 - P(x<3282) = 0.15866

Converting to percentage

0.15866 × 100 = 15.866%

The percent of flights on these particular routes that will burn more than 3,282 gallons of kerosene is 15.866%

3 0
3 years ago
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