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2 years ago

1/2(4x-2)-2/3(6x+9<4

1 answer:

6 0

$\frac{1}{2}$ (4x - 2) - $\frac{2}{3}$ (6x + 9) < 4 Use the Distributive Property
2x - 1 - 4x + 6 < 4 Combine like terms (-1 and 6)
2x + 5 - 4x < 4 Combine (2x and -4x)
-2x + 5 < 4 Subtract 5 from both sides
-2x < -1 Divide both sides by -2, don't forget to flip the inequality sign after dividing by a negative
x > $\frac{1}{2}$

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5(2)>3(1)+4

10 > 7. Is True because 10 is greather than 7. All needed to do is evaluate the x and y variables for each possibility. A, B, C D. And see if is a logical answer.

Any other combination gives you a false statement.

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3 years ago

Help simplify please choose the correct answer~!!!!!

ANEK [815]

Answer:

Step-by-step explanation:

Given

-4x²(5 $x^{4}$ - 3x² + x - 2)

Multiply each term in the parenthesis by - 4x²

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3 years ago

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nevsk [136]

Answer:

(0,-2)

(2,-1)

(4,0)

Step-by-step explanation:

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2 years ago

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If a person rolls a six dash sided dierolls a six-sided die and then flips a coinflips a coin, describe the sample space of pos

Olegator [25]

Answer:

S={1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}

Step-by-step explanation:

The sample space for a six sided die is {1,2,3,4,5,6} as there are 6 possible outcomes and for flipping a coin is {H,T} as there are two possible outcomes.

If a person rolls a six sided die and then flips a coin then the sample space for this event will be written as

S={1H,2H,3H,4H,5H,6H,1T,2T,3T,4T,5T,6T}.

The number of elements in the sample space are

n(S)=12.

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2 years ago

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Construct a 90% confidence interval for μ1-μ2 with the sample statistics for mean calorie content of two bakeries' specialty pi

DIA [1.3K]

Answer:

The 90% confidence interval for the difference in mean (μ₁ - μ₂) for the two bakeries is; (49) < μ₁ - μ₂ < (289)

Step-by-step explanation:

The given data are;

Bakery A

$\overline x_1$ = 1,880 cal

s₁ = 148 cal

n₁ = 10

Bakery B

$\overline x_2$ = 1,711 cal

s₂ = 192 cal

n₂ = 10

$\left (\bar{x}_1-\bar{x}_{2} \right ) - t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}< \mu _{1}-\mu _{2}< \left (\bar{x}_1-\bar{x}_{2} \right ) + t_{c}\cdot \hat \sigma \sqrt{\dfrac{1}{n_{1}}+\dfrac{1}{n_{2}}}$

df = n₁ + n₂ - 2

∴ df = 10 + 18 - 2 = 26

From the t-table, we have, for two tails, $t_c$ = 1.706

$\hat{\sigma} =\sqrt{\dfrac{\left ( n_{1}-1 \right )\cdot s_{1}^{2} +\left ( n_{2}-1 \right )\cdot s_{2}^{2}}{n_{1}+n_{2}-2}}$

$\hat{\sigma} =\sqrt{\dfrac{\left ( 10-1 \right )\cdot 148^{2} +\left ( 18-1 \right )\cdot 192^{2}}{10+18-2}}= 178.004321469$

$\hat \sigma$ ≈ 178

Therefore, we get;

$\left (1,880-1,711 \right ) - 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}< \mu _{1}-\mu _{2}< \left (1,880-1,711 \right ) + 1.706\times178 \sqrt{\dfrac{1}{10}+\dfrac{1}{18}}$

Which gives;

$169 - \dfrac{75917\cdot \sqrt{35} }{3,750} < \mu _{1}-\mu _{2}< 169 + \dfrac{75917\cdot \sqrt{35} }{3,750}$

Therefore, by rounding to the nearest integer, we have;

The 90% C.I. ≈ 49 < μ₁ - μ₂ < 289

4 0

3 years ago