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3 years ago

WILL

Mathematics

2 answers:

IRISSAK [1]3 years ago

6 0

Answer:

-10, -|-8|, -(-2), |-11|, 5²

Step-by-step explanation:

Juli2301 [7.4K]3 years ago

4 0

Answer:

YOU POST THIS TWICE

IT'S -10, -|-8|, -(-2), |-11|, 5²

Step-by-step explanation:

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Which is the definition of perpendicular lines that includes only necessary and sufficient information?

Papessa [141]

Two lines that intersect to create a 90 degree angle

6 0

3 years ago

Move two choices to the blanks to correctly complete the sentences.

Pavel [41]

Answer:

(A) There should have been 5 outcomes of HT

(B) The experimental probability is greater than the theoretical probability of HT.

Step-by-step explanation:

Given

$S = \{HH,HT,TH,TT\}$ -- Sample Space

$n(S) = 4$ --- Sample Size

Solving (a); theoretical outcome of HT in 20 tosses

First, calculate the theoretical probability of HT

$P(HT) = \frac{n(HT)}{n(S)}$

$P(HT) = \frac{1}{4}$

Multiply this by the number of tosses

$P(HT) * n= \frac{1}{4} * 20$

$P(HT) * n= 5$

Solving (b); experimental probability of HT

Here, we make use of the table

$P(HT) = \frac{n(HT)}{n(S)}$

$P(HT) = \frac{6}{20}$

$P(HT) = 0.30$ ---- Experimental Probability

In (a), the theoretical probability is:

$P(HT) = \frac{1}{4}$

$P(HT) = 0.25$ ---- Experimental Probability

By comparison;

$0.30 > 0.25$

6 0

3 years ago

Please I need help. I need this ASAP, the screenshot is right below here.

Delicious77 [7]

Answer:

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Step-by-step explanation:

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6 0

3 years ago

Factor: 81y^2-49<br> Your answer

oksian1 [2.3K]

Answer:(9y-7)(9y+7)

Step-by-step explanation:

81y^2-49

=(9y)^2-(7)^2

=(9y-7)(9y+7)

4 0

2 years ago

Factor the polynomial function over the complex numbers. f(x)=x^4-x^3-2x−4 Enter your answer in the box. f(x) = The answer is: (

Vladimir79 [104]

Answer:

Factors: $(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

Step-by-step explanation:

We are given a polynomial:

$f(x)=x^4-x^3-2x-4$

We have to factor the given polynomial into its complex factors.

The factorization can be done as follows:

$f(x)=x^4-x^3-2x-4 = 0\\x^4-x^3-2x-4 = 0\\x^4-4-x^3-2x=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^4-4)-(x^3+2x) = 0\\(x^2+2)(x^2-2)-x(x^2+2) = 0\\(x^2+2)(x^2-2-x) = 0\\(x^2+2)(x^2-x-2) = 0\\(x^2+2)(x^2-2x-+x-2) = 0\\(x^2+2)((x(x-2)+1(x-2))=0\\(x^2+2)(x+1)(x-2)=0\\\text{Identity: }a^2-b^2 = (a+b)(a-b)\\(x^2-(\sqrt{-2})^2)(x+1)(x-2)=0\\(x+i\sqrt2)(x-i\sqrt2)(x+1)(x-2)=0$

8 0

2 years ago