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Gnoma [55]
2 years ago
15

Two stores sell the same shirt at the same price. Both stores decide to sell the shirt at a discounted price.

Mathematics
1 answer:
Liula [17]2 years ago
4 0
A option is the correct one 1 is the correct one by the number
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The actual answere 3 + 1(6 ÷ 2)
nadezda [96]

Hey there!

Just do PEMDAS

• Parentheses

• Exponents

• Multiplication

• Division

• Addition

• Subtraction

3 + 1(6 ÷ 2)

= 3 + 1(3)

= 3 + 3

= 6

Therefore, your answer is: 6

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

4 0
2 years ago
What are the steps for solving, 7,542÷3
valina [46]

Answer:

2514

Step-by-step explanation:

Do the long division and you'll find the answer.

8 0
3 years ago
Keegan is priting and selling his original design on t-shirts. He has concluded that for x shirts, in thousands sold his total p
Damm [24]

Complete question is;

Keegan is printing and selling his original design on t-shirts. He has concluded that for x shirts, in thousands sold his total profits will be p(x) = -x³ + 4x² + x dollars, in thousands will be earned. How many t-shirts (rounded to the nearest whole number) should he print in order to make maximum profits? What will his profits rounded to the nearest whole dollar be if he prints that number of shirts?

Answer:

Number of t-shirts to make maximum profit = 2790 shirts

Maximum profit = $12,209

Step-by-step explanation:

From the question, we are given that the profit function is;

p(x) = -x³ + 4x² + x

For the maximum value of the profit function,

(dp/dx) = 0 and (d²p/dx²) < 0

Since, p(x) = -x³ + 4x² + x

Then,

(dp/dx) = -3x² + 8x + 1

at maximum point (dp/dx) = 0, thus;

-3x² + 8x + 1 = 0

Solving this using quadratic formula, the roots are;

x = -0.12 or 2.79

Also, (d²p/dx²) = -6x + 8

Now, let's put the roots of x into -6x + 8 and check for maximum value conditon;

at x = -0.12

(d²p/dx²) = -6(0.12) + 8 = 7.28 > 0

At x = 2.79

(d²p/dx²) = -6(2.79) + 8 = -8.74 < 0

Maximum has to be d²p/dx² < 0

So, the one that meets the condition is -8.74 < 0 at x = 2.79

Thus, the maximum of the profit function exists when the number of shirts, x = 2.79 (in thousands) = 2790

Now, the maximum profits that corresponds to this number of t-shirts of 2.79(in thousands) is obtained by putting 2.79 for x in the profit function;

So,

p(2.79) = -(2.79)³ + 4(2.79²) + 2.79

p(x) = -21.7176 + 31.1364 + 2.79

p(x) = 12.2088 (in thousand dollars) ≈ $12,209

6 0
3 years ago
The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population o
blondinia [14]

The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

According to the statement

we have to find that the standard deviation, mean and the intervals from the given data.

So, According to the given data from the method of tree ring dating

The value of mean is

x-bar = (1271 + 1208 + 1229 + 1299 + 1268 + 1316 + 1275 + 1317 + 1275) / 9 = x-bar = 1273 AD

And now we find standard deviation :

s = √∑(xi - x-bar) / (N - 1)

∑(xi - x-bar)^2 = (1271 - 1273)2 + (1208 - 1273)2 + (1229 - 1273)2 + ... + (1275 - 1273)2

∑(xi - x-bar)^2 = (-2)2 + (-65)2 + (-44)2 + ... + (2)2

∑(xi - x-bar)^2 = 4 + 4225 + 1936 + 676 + 25 + 1849 + 4 1936 + 4

∑(xi - x-bar)^2 = 10,659

Now,

s^2 = 10659/8 = 1332

s = 37 years

So, standard deviation is 37 years.

We need the t-distribution table since the standard deviation is unknown.  Therefore, our degrees of freedom is 9 - 1 = 8 and the critical value is 1.860.  Set up the confidence interval for the mean:

[x-bar ± t*(s/√n)] = [1273 ± 1.860*(37/√9)]

[x-bar ± t*(s/√n)] = [1250,1296]

So, The value of mean, standard deviation and interval from the method of tree ring dating is 1273 AD, 37 years, [1250,1296].

Learn more about method of tree ring dating here

brainly.com/question/15107034

Disclaimer: This question was incomplete. Please find the full content below.

Question:

The method of tree ring dating gave the following years A.D. for an archaeological excavation site. Assume that the population of x values has an approximately normal distribution.

For more data please see the image below.

#SPJ4

4 0
2 years ago
Jenna, Tatiana, and Nina have decided to share their Halloween candy.
Shtirlitz [24]
Idkgccghhhbbjjfxgvbijv
7 0
3 years ago
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