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3 years ago

Find the surface area of the above solid. A. 52 cm2 B. 43 cm2 C. 54 cm2 D. 58 cm2

Mathematics

1 answer:

PSYCHO15rus [73]3 years ago

6 0

Answer:

A.52 cm2 ihope ihelp po hehe

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If you place $1,000.00 in a savings account with an interest rate of 5.25%/month, how much should you earn in interest at the en

Art [367]

Interest earned=principle×interest rate×time
I=prt
I=1,000×0.0525×1
I=52.5

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4 years ago

Can anyone help me solve a trigonomic identity problem and also help me how to do it step by step?

dusya [7]

$\bf cot(\theta)=\cfrac{cos(\theta)}{sin(\theta)}
\qquad csc(\theta)=\cfrac{1}{sin(\theta)}
\\\\\\
sin^2(\theta)+cos^2(\theta)=1\\\\
-------------------------------\\\\$

$\bf \cfrac{cos(\theta )cot(\theta )}{1-sin(\theta )}-1=csc(\theta )\\\\
-------------------------------\\\\
\cfrac{cos(\theta )\cdot \frac{cos(\theta )}{sin(\theta )}}{1-sin(\theta )}-1\implies \cfrac{\frac{cos^2(\theta )}{sin(\theta )}}{\frac{1-sin(\theta )}{1}}-1\implies 
\cfrac{cos^2(\theta )}{sin(\theta )}\cdot \cfrac{1}{1-sin(\theta )}-1
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\cfrac{cos^2(\theta )-1[sin(\theta )[1-sin(\theta )]]}{sin(\theta )[1-sin(\theta )]}$

$\bf \cfrac{cos^2(\theta )-1[sin(\theta )-sin^2(\theta )]}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{cos^2(\theta )-sin(\theta )+sin^2(\theta )}{sin(\theta )[1-sin(\theta )]}
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\cfrac{cos^2(\theta )+sin^2(\theta )-sin(\theta )}{sin(\theta )[1-sin(\theta )]}\implies \cfrac{\underline{1-sin(\theta )}}{sin(\theta )\underline{[1-sin(\theta )]}}
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7 0

3 years ago

Write $1.50 to $2.50 as a fraction in simplest form

Andre45 [30]

1/2, I divided 2.50 by 1.50 and ended up getting 1/2.

8 0

3 years ago

Use the graph that shows the solution to f(x)=g(x) .

n200080 [17]

Answer:

1. x=3 x=1

2. x=0

Step-by-step explanation:

We want f(x) to equal g(x)

f(x)=g(x)

1/(x-2) = (x-2)

Using cross products

1 = (x-2) * (x-2)

1 = x^2 -2x-2x+4

Subtract 1 from each side

1-1 = ^2 -2x-2x+4-1

0 = x^2 -4x+3

Factoring

What number multiplies to 3 and adds to -4

-3*-1 = 3

-3+-1 = -4

0= (x-3) (x-1)

Using the zero product property

x-3 =0 x-1=0

x=3 x=1

2. f(x)=x^2+4x+2

g(x)=(1/2)^ x+1

From the graph,we can see they intersect at x=0

5 0

4 years ago

All athletes at the Olympic Games (OG) are tested for performance-enhancing steroid drug use. The basic Anabolic Steroid Test (A

Semenov [28]

Answer:

a ) the probability of using steroids and having a negative test is 0.5%

b) The probability of testing positive is 6.4%

c) The probability of not using steroids, given that the test is negative is 99.47%

d) No, they are not statistically indepent.

e) The probability that the athlete will either use steroids or test positive is 6.9%

Step-by-step explanation:

Let A be the event that the test result is positive and B the event that the athlete uses Steroids. We are given the following

$P(A|B) = 90%, P(A|B^c) = 2%, P(B) = 5%$

From which we deduce that

$P(A^c|B) = 10%, P(B^c) = 95%$

a) We are asked for the probability $P(A^c\cap B)$ . REcall the conditional probability formula that, given two events C,D the conditional probability $P(C|D) = \frac{P(C\cap D)}{P(D)}$ . Then we have that

$P(A^c\cap B) = P(A^c|B)P(B) = 10\% \cdot 5\%=0.5\%$ .

b) We are asked for the probability P(A). We can use the fact that given two mutually exclusive events(that is, whose intersection is empty) A,B the probability P(C) of an event is given by $P(C) = P(C|A)P(A)+P(C|B)P(B)$ . Then

$P(A) = P(A|B)P(B)+P(A|B^c)P(B^c) = 90\%\cdot5\% + 2\% \cdot 95% = 6.4\%$

c) We are asked for the probability P(B^c|A^c). Recall that $P(A|B) = \frac{P(B|A)P(A)}{P(B)}$ . Then

$P(B^c|A^c) = \frac{P(A^c|B^c)P(B^c)}{P(A^c)}= \frac{P(A^c|B^c)P(B^c)}{1-P(A)}= \frac{(1-P(A|B^c))P(B^c)}{1-P(A)}=\frac{98\%\cdot 95\%}{1-6.4\%}= 99.47\%$

d) We say that two events A,B are statistically indepent if P(A|B) = P(A). Note that from point B the probability of testing negative is 1- 6.4% = 93.6%. Since 93.6% is different from 99.47% this means that testing positive and using steroids are not statistically independent.

e) We are asked for the probability $P(A\cup B)$ . We use the following

$P(A\cupB) = P(A)+P(B)-P(A\cap B) = P(A) +P(B)-P(A|B)P(B) = 6.4\%+5\%-90\%\cdot 5\%=6.9\%$

4 0

4 years ago