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2 years ago

Solve: |x+2|−1≥5. Write your solution in interval notation.

Mathematics

1 answer:

AlekseyPX2 years ago

6 0

nswer:

Step-by-step explanat

|x+2| - 1 ≥ 5

 |x+2| ≥ 5+1

 |x+2| ≥ 6

 x+2 ≥ 6 hoặc x+2 ≤ -6

+ với x+2 ≥ 6  x ≥ 6 – 2  x ≥ 4

+ với x+2 ≤ -6  x ≤ -6 – 2  x ≤ -8

(-∝;-8)∪(4;+∝)

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The scores of high school seniors on the act college entrance examination in a recent year had mean Î¼ = 20.8 and standard devia

natka813 [3]

To solve this problem, we make use of the z statistic. The formula for z score is:

z = (x – μ) / s

where,

x = is the sample score = 21 or higher

μ = the population mean = 20.8

s = the standard deviation = 4.8

Solving for the z score:

z = (21 – 20.8) / 4.8

z = 0.0417

Next step to do is to find for the p value using the standard distribution tables at z = 0.04. Since we are looking for the probability of getting 21 or higher, therefore this is a right tailed test, hence

p = 0.516

<span>So there is about 51.6% that a student will get a score of 21 or greater.</span>

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2 years ago

Evaluate the following expression 1/3^2

NISA [10]

Answer:

1/9

Step-by-step explanation:

1/3*1/3=1/9

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3 years ago

PART A: Mrs. konsdorf claims that angle R is a right angle.Is Mrs. konsdorf correct? explain your reasoning PART B: if T is trab

Anna007 [38]

Answer:

Part A: Angle R is not a right angle.

Part B; Angle GRT' is a right angle.

Step-by-step explanation:

Part A:

From the given figure it is noticed that the vertices of the triangle are G(-6,5), R(-3,1) and T(2,6).

Slope formula

$m=\frac{y_2-y_1}{x_2-x_1}$

The product of slopes of two perpendicular lines is -1.

Slope of GR is

$\text{Slope of GR}=\frac{1-5}{-3-(-6)}=\frac{-4}{3}$

Slope of RT is

$\text{Slope of RT}=\frac{6-1}{2-(-3)}=\frac{5}{5}=1$

Product of slopes of GR and RT is

$\frac{-4}{3}\times 1=\frac{-4}{3}\neq -1$

Therefore lines GR and RT are not perpendicular to each other and angle R is not a right angle.

Part B:

If vertex T translated by rule

$(x,y)\rightarrow(x-1,y-2)$

Then the coordinates of T' are

$(2,6)\rightarrow(2-1,6-2)$

$(2,6)\rightarrow(1,4)$

Slope of RT' is

$\text{Slope of RT'}=\frac{4-1}{1-(-3)}=\frac{3}{4}$

Product of slopes of GR and RT' is

$\frac{-4}{3}\times \frac{3}{4}=-1$

Since the product of slopes is -1, therefore the lines GR and RT' are perpendicular to each other and angle GRT' is a right angle.

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2 years ago

Solve the equation for y (y-3)^2=4y-12

Ket [755]

(y -3)^2 = 4y - 12

Expand.

(y-3)(y -3) = 4y - 12

y² - 6y + 9 = 4y - 12

y² - 6y -4y + 9 + 12 = 0

y² - 10y + 21 = 0

Factors are -7y and - 3y

y² - 10y + 21 = 0

y² - 3y - 7y + 21 = 0

y(y - 3) -7(y - 3) = 0

(y - 3)(y - 7) = 0

y-3 = 0 or y - 7 = 0

y = 0 + 3 or y = 0 + 7

y = 3 or 7

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2 years ago

Divide by 12 nd equal 14

ICE Princess25 [194]

170 dived by 12 is 14

7 0

2 years ago

Read 2 more answers