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3 years ago

(3x-8)° (4x-23)° fine the x

Mathematics

1 answer:

Veronika [31]3 years ago

7 0

You cannot find x.

Because both phrases are to the power of 0, we know that it is essentially 1*1

So:

(3x - 8)^0 (4x - 23)^0 = 1

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The quality control manager of a soda manufacturing company wishes to use a significant level of 0.05 to test whether the variab

Alex17521 [72]

Answer: 1.645

Step-by-step explanation:

Given : The quality control manager of a soda manufacturing company wishes to use a significant level of 0.05 to test whether the variabilities in the amount of soda in the company's 16 OZ bottles is more than the variabilities in the company's 12 OZ bottles.

Let $\sigma_1^2$ represents the variance for 16 OZ bottles and $\sigma_2^2$ represents the variance for 12 OZ bottles.

The Null and Alternative Hypothesis will be :-

$H_0:\sigma_1^2\geq\sigma_2^2$

$H_1:\sigma_1^2$ , since alternative hypothesis is left-tailed , then the test is left-tailed test.

By using the standard normal distribution table for z, the critical value for this hypothesis:-

$z_{\alpha}=z_{0.05}=1.645$

6 0

4 years ago

Read 2 more answers

2.240<br> F = d + 2.24<br> What is the answer

ExtremeBDS [4]

Answer:

F= 0.44642857d+1

Step-by-step explanation:

4 0

3 years ago

Billy and Joe go to McDonalds for lunch. They order 2 burgers, 3 fries, and 2 sodas. The mom

julia-pushkina [17]

Answer:

2.43 and rounded it is 2 dollars

Step-by-step explanation:

2 burger-$7.00

3 Fries-

2 Sodas- $2.00

7+2=9

$16.29-9.00=7.29

$7.29/3=2.43

5 0

3 years ago

An engineer is designing a large steel pad to be installed on the deck of an aircraft carrier. Its total volume will be 18 yd^3.

Shtirlitz [24]

Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0. $\overline 1$ yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0. $\overline 1$ yd. = (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel pad, m₁ = ρ × V₁

∴ The mass of the steel pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

∴ C = 4,050 lbs × $55/lb = $222,750

6 0

3 years ago

Let n be a natural number. Show that 3 | n3 −n

Vladimir [108]

Let $n=1$ . Then $n^3-n=1^3-1=0$ . By convention, every non-zero integer $n$ divides 0, so $3\vert n^3-n$ .

Suppose this relation holds for $n=k$ , i.e. $3\vert k^3-k$ . We then hope to show it must also hold for $n=k+1$ .

You have

$(k+1)^3-(k+1)=(k^3+3k^2+3k+1)-(k+1)=(k^3-k)+3(k^2+k)$

We assumed that $3\vert k^3-k$ , and it's clear that $3\vert 3(k^2+k)$ because $3(k^2+k)$ is a multiple of 3. This means the remainder upon divides $(k+1)^3-(k+1)$ must be 0, and therefore the relation holds for $n=k+1$ . This proves the statement.

4 0

3 years ago