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3 years ago

Yo 10 or 22 points pls help I rlly need help !!!

Mathematics

1 answer:

NISA [10]3 years ago

8 0

3x < -21
x < -7
And then shade to the left 7 places.

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Help pls i have one chance left

emmasim [6.3K]

Answer:

2^8x-3

Step-by-step explanation:

5 0

3 years ago

Need help with statistics plssssss

podryga [215]

Answer:

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5 0

3 years ago

Read 2 more answers

Please someone answer this please

gladu [14]

Option C:

$\frac{3 a^{2} b^{11}}{2 }$ is equivalent to the given expression.

Solution:

Given expression:

$$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}$

To find which expression is equivalent to the given expression.

$$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}$

Using exponent rule: $\frac{1}{a^m}=a^{-m}, \ \ \frac{1}{a^{-m}}=a^{m}$

$$=\frac{-18 a^{-2} b^{5}a^{4} b^{6}}{-12 }$

$$=\frac{-18 a^{-2} a^{4} b^{5} b^{6}}{-12 }$

Using exponent rule: ${a^m}\cdot{a^n}=a^{m+n}$

$$=\frac{-18 a^{(-2+4)} b^{(5+6)}}{-12 }$

$$=\frac{-18 a^{2} b^{11}}{-12 }$

Divide both numerator and denominator by the common factor –6.

$$=\frac{3 a^{2} b^{11}}{2 }$

$$\frac{-18 a^{-2} b^{5}}{-12 a^{-4} b^{-6}}=\frac{3 a^{2} b^{11}}{2 }$

Therefore, $\frac{3 a^{2} b^{11}}{2 }$ is equivalent to the given expression.

Hence Option C is the correct answer.

8 0

3 years ago

2x + 5<br> Find the val<br> 3x+15

Crank

Answer:

2x + 5: Solve

$2x + 5 - 5 = 0 - 5$

$2x / 2 = -5 / 2$

$x = -5/2$

3x + 15: Solve

$3x + 15 - 15 = 0 - 15$

$3x / 3 = -15 / 3$

$x = -5$

3 0

3 years ago

A fair coin is tossed three times and the events A, B, and C are defined as follows: A:{ At least one head is observed } B:{ At

kicyunya [14]

Answer:

(a) 1/2

(b) 1/2

(c) 1/8

Step-by-step explanation:

Since, when a fair coin is tossed three times,

The the total number of possible outcomes

n(S) = 2 × 2 × 2

= 8 { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT },

Here, B : { At least two heads are observed } ,

⇒ B = {HHH, HHT, HTH, THH},

⇒ n(B) = 4,

Since,

$\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}$

(a) So, the probability of B,

$P(B) =\frac{n(B)}{n(S)}=\frac{4}{8}=\frac{1}{2}$

(b) A : { At least one head is observed },

⇒ A = {HHH, HHT, HTH, THH, HTT, THT, TTH},

∵ A ∩ B = {HHH, HHT, HTH, THH},

n(A∩ B) = 4,

$\implies P(A\cap B) = \frac{n(A\cap B)}{n(S)} = \frac{4}{8}=\frac{1}{2}$

(c) C: { The number of heads observed is odd },

⇒ C = { HHH, HTT, THT, TTH},

∵ A ∩ B ∩ C = {HHH},

⇒ n(A ∩ B ∩ C) = 1,

$\implies P(A\cap B\cap C)=\frac{1}{8}$

7 0

3 years ago