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Ierofanga [76]
2 years ago
13

Sammy wrote the following proof. What did he do wrong?

Mathematics
2 answers:
seraphim [82]2 years ago
8 0

i think it would be D i hope i helped

OLEGan [10]2 years ago
3 0

Answer:

Choice B is correct answer.

Step-by-step explanation:

We have given a expression.

cos²∅tan²∅

We have to simplify above expression.

Since, we know that

tan²∅ +1 = sec²∅

cos²∅tan²∅ = cos²∅(sec²∅-1)

cos²∅tan²∅ = cos²∅sec²∅-cos²∅

since, sec∅ is reciprocal of cos∅.

sec∅ = 1/cos∅

cos²∅tan²∅ = cos²∅(1/cos²∅)-cos²∅

cos²∅tan²∅ = 1-cos²∅

Since, from following equation;

cos²∅+sin²∅ = 1

cos²∅tan²∅ = sin²∅

Hence, choice B is the answer.

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Olivia wants to cut 3 3/4 inches from a piece of string.She has already cut off 2 9/16 inches from the piece of string.How much
Roman55 [17]

Answer: she still has to cut off 19/16 inches.

Step-by-step explanation:

Olivia wants to cut 3 3/4 inches from a piece of string. Converting 3 3/4 inches to improper fraction, it becomes 15/4 inches.

She has already cut off 2 9/16 inches from the piece of string. Converting 2 9/16 inches to improper fraction, it becomes 41/16 inches.

Therefore, the length of string that is left for her to cut off would be

15/4 - 41/16 = (60 - 41)/16

= 19/16 inches

6 0
2 years ago
In a multiple choice exam, there are 5 questions and 4 choices for each question (a, b, c, d). Nancy has not studied for the exa
KonstantinChe [14]

Answer:

Part 1

a) 0.0791

b) 0.000977

c) 0.7627

Part 2

a) 0.049

b) 0.067

c) 0.864 or 0.136 (depending on what the question truly says)

d) 0.806

Step-by-step explanation:

Part 1

Since there are 4 choices per question, and only one correct answer per question.

The probability of getting a question right = (1/4) = 0.25

Probability of getting a question wrong = 1 - 0.25 = 0.75

a) Probability that the first question she gets right is the 5th question means she gets the first 4 questions wrong, and gets the last question.

0.75 × 0.75 × 0.75 × 0.75 × 0.25 = 0.0791

b) Probability that she gets all of the questions right

0.25 × 0.25 × 0.25 × 0.25 × 0.25 = 0.000977

c) Probability that she gets at least one question right = 1 - (probability that she doesn't get any question right) = 1 - (0.75⁵) = 1 - 0.2373 = 0.7627

Part 2

We use standard normal distribution for this

a) Area under (Z< -1.65) = P(z < -1.65) = 1 - P(z ≥ -1.65) = 1 - P(z ≤ 1.65) = 1 - 0.951 = 0.049

b) Area under (Z > 1.5) = P(z > 1.5) = 1 - P(z ≤ 1.5) = 1 - 0.933 = 0.067

c) P(z > -1.1) or P(z < -1.1)

P(z > - 1.1) = 1 - P(z ≤ -1.1) = 1 - 0.136 = 0.864

P(z < - 1.1) = 1 - P(z ≥ - 1.1) = 1 - P(z ≤ 1.1) = 1 - 0.864 = 0.136

d) |Z|>1.3 = P(-1.3 < z < 1.3) = P(z < 1.3) - P(z < -1.3)

P(z < 1.3) = 1 - P(z ≥ 1.3) = 1 - P(z ≤ -1.3) = 1 - 0.097 = 0.903

P(z < -1.3) = 1 - P(z ≥ -1.3) = 1 - P(z ≤ 1.3) = 1 - 0.903 = 0.097

P(z < 1.3) - P(z < -1.3) = 0.903 - 0.097 = 0.806

6 0
2 years ago
Given: ΔABC, AC = BC, AB = 3 CD ⊥ AB, CD = √3 Find: AC
valina [46]

Answer:

\boxed{AC = 2.3}

Step-by-step explanation:

AD = BD  (CD bisects AB means that it divides the line into two equal parts)

So,

AD = BD = AB/2

So,

AD = 3/2

AD = 1.5

<u><em>Now, Finding AC using Pythagorean Theorem:</em></u>

c^2 = a^2+b^2

Where c is hypotenuse (AC), a is base (AD) and b is perpendicular (CD)

AC^2= (1.5)^2+(\sqrt{3} )^2

AC^2 = 2.25 + 3

AC^2 = 5.25

Taking sqrt on both sides

AC = 2.3

3 0
1 year ago
Read 2 more answers
bookstore ordered 1,528 packs of pens and 1,823 packs of pencils at the prices shown. how much did the bookstore spend on pens?
IgorC [24]
I found that the given prices are $ 4 per pack of pens and $ 3 per pack of penciles.

The question is how much the bookstore spent on pens.

Then you have to multiply the number packs of pens, which is 1,528, times the price per pack pens which is $ 4.

So, the answer is: 1528 packs of pens * $ 4 / pack of pens = $6112.


8 0
2 years ago
For the lunch special at a high school cafeteria, students can get either salad or french fries as a side order. The following t
LekaFEV [45]

Answer: E

Step-by-step explanation:

8 0
1 year ago
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