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andrey2020 [161]
3 years ago
5

Which number line represents the solution set for the inequality -4(x + 3) 3-2-2x? -7 -6 -5 -4 -3 -2 -1 0 1 G 4 4 4 6 OT 7 -7 -6

-5 -4 -3 -2 -1 3 4 091 6 7 tuum ແຕ່ແຕ່nານານແນານານານານາສາສນາພແ[ -7 -6 -5 -4 -3 -2 -1 0 1 B 5 5 6 T -7 -6 -5 -4 -3 -2 -1 0 1 3 4 A
HURRY!!!!​

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
8 0

Answer:

Option (1)

Step-by-step explanation:

Given inequality is,

-4(x + 3) ≤ -2 - 2x

Further solve this inequality,

-4x - 12 ≤ -2 - 2x

-4x + 2x - 12 ≤ -2 - 2x + 2x

-2x - 12 ≤ -2

-2x - 12 + 12 ≤ -2 + 12

-2x ≤ 10

-x ≤ 5

x ≥ -5

Therefore, on a number line draw an arrow starting with a solid circle (because of equal to sign) and moving towards the numbers greater than -5.

Option (1) is the answer.

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Adult = $9.80

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m_a_m_a [10]
If you cannot read it, let me know.

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3 years ago
In rectangle abcd, points p and q lie on side AB and DC respectively. Angle PMQ is a right angle, M is the midpoint of side BC a
nirvana33 [79]

Answer:

PM:MQ = 8:3.

Step-by-step explanation:

\rm \angle B\hat{M}P + 90^{\circ} + \angle C\hat{M}Q = 180^{\circ};

\implies \rm \angle B\hat{M}P + \angle C\hat{M}Q = 90^{\circ};

\implies \rm 90^{\circ} - \angle B\hat{M}P = \angle C\hat{M}Q.

Also,

\rm \angle B\hat{P}M = 90^{\circ} - \angle B\hat{M}P in right triangle PBM.

Thus \rm \angle{P\hat{B}M} = \angle C\hat{M}Q.

Additionally \rm \angle \hat{B} = 90^{\circ} = \angle \hat{C}.

Therefore \rm \triangle PBM \sim \triangle MCQ.

\rm \displaystyle BC = 2\;MC for M is the midpoint of segment BC.

\rm \displaystyle PB = \frac{4}{3}BC = \frac{8}{3}MC.

\rm \triangle PBM \sim \triangle MCQ implies that

\displaystyle \rm PM:MQ = PB:MC = 1:\frac{8}{3} = 8:3.

8 0
2 years ago
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DedPeter [7]

Answer:

t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }

Step-by-step explanation:

H0: µ1 – µ2 = 0

HA: µ1 – µ2 ≠ 0

We have given,

The population variances are not known and cannot be assumed equal.

The test statistic for the test is

t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }

Where,

\bar x_1 = sample meaan of population 1

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n_1 = sample size of population 1

n_2 = sample size of population 2

Therefore, this is the test

t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }

7 0
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