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3 years ago

Please help me with this math question!

Mathematics

2 answers:

NARA [144]3 years ago

6 0

Option A would be the correct one difference/product :)

sergiy2304 [10]3 years ago

5 0

The answer is A have a good night

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What arithmetic variables look like

juin [17]

Answer:

LETTERS

Step-by-step explanation:

In Arithmetic, variables look like LETTERS.

5 0

3 years ago

Help. Please. Thanks.

Musya8 [376]

950 * .05 = $47.5
550 * .04 = $22

$47.5 + $22 = $69.50 interest earned

answer: amount in 4% account = $550

6 0

3 years ago

What is the radius of the base of the cone in the figure below?

liubo4ka [24]

Answer: approximately 12.17 units

==========================================================

Work Shown:

We'll use the tangent ratio to find r

tan(angle) = opposite/adjacent

tan(49) = 14/r

r*tan(49) = 14

r = 14/tan(49)

r = 12.1700143294271

r = 12.17

The radius is approximately 12.17 units long.

7 0

2 years ago

Eleanor scores 680 on the mathematics part of the SAT. The distribution of SAT math scores in recent years has been Normal with

Katen [24]

Answer:

Elanor's standardized score is 1.19
Gerald's standardized score is 0.72
Elanor has higher score

Step-by-step explanation:

To compare Elanor's and Gerald's math scores, we need to standardize them and calculate their z-scores.

z score can be calculated using the formula

z= $\frac{X-M}{s}$ where

X is the student's score

M is the mean score of the exam

s is the standard deviation of the exam

Elanor's standardized score is:

z(e) = $\frac{680-554}{106}$ ≈ 1.19

Gerald's standardized score is:

z(g)= $\frac{27-24.7}{3.2}$ ≈ 0.72

Since z(e) > z(g), Elanor has higher score

6 0

3 years ago

A coin, having probability p of landing heads, is continually flipped until at least one head and one tail have been flipped. (a

Natali [406]

Answer:

(a)

The probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

Step-by-step explanation:

(a)

Case 1

Imagine that you throw your coin and you get only heads, then you would stop when you get the first tail. So the probability that you stop at the fifth flip would be

$p^4 (1-p)$

Case 2

Imagine that you throw your coin and you get only tails, then you would stop when you get the first head. So the probability that you stop at the fifth flip would be

$(1-p)^4p$

Therefore the probability that you stop at the fifth flip would be

$p^4 (1-p) + (1-p)^4 p$

(b)

The expected numbers of flips needed would be

$\sum\limits_{n=1}^{\infty} n p(1-p)^{n-1} = 1/p$

Therefore, suppose that $p = 0.5$ , then the expected number of flips needed would be 1/0.5 = 2.

7 0

3 years ago