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Semmy [17]
2 years ago
12

Hello please help i’ll give brainliest

Mathematics
1 answer:
Alborosie2 years ago
3 0

Answer:

C

or "It led to the understanding of hieroglyphics"

Step-by-step explanation:

It's the Rosetta Stone artifact.

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Assume that two marbles are drawn without replacement from a box with 1 blue, 3 white, 2 green and 2 red marbles. Find probabili
aniked [119]
Please answer please please thank you
8 0
3 years ago
Two sides of a right triangle measure 2 units and 4 units.
Liono4ka [1.6K]

Step-by-step explanation:

a/q first we have to find the measurement of third side

so,

h² = a² + b²

4² = a² + 2²

16 = a² + 4

a = 16 - 4 = 12

a² = 12

a = √12 = 3.46

so,

area of square = a × a

= 3.46 × 3.46 = 11.97 sq unit

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8 0
3 years ago
AREA
natka813 [3]

Answer:

Show the figure

Step-by-step explanation:

3 0
3 years ago
Solve. (x-4)²=5<br> PLEASE HELP
goldenfox [79]

Answer:

A

Step-by-step explanation:

(x-4)^2=5

(x-4)(x-4)=5

x(x-4)-4(x-4)=5

distribute

x^2-4x-4x+16=5

combine like terms

x^2-8x+16=5

move terms to the left and combine

x^2-8x+11=0

use the quadratic formula and simplify

x=-(-8)+/-2 to the square roots of 5 over 2

separate and solve

x=4+ square root of 5

x=4- square root of 5

ANSWER: x=4+/- square root of 5

8 0
2 years ago
A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the prob
UkoKoshka [18]

Answer:

The probability that 8 mice are​ required is 0.2428.

Step-by-step explanation:

Given : A scientist inoculates​ mice, one at a​ time, with a disease germ until he finds 3 that have contracted the disease. If the probability of contracting the disease is two sevenths.

To find : What is the probability that 8 mice are​ required? The probability that that 8 mice are required is nothing ?

Solution :

Applying binomial distribution,

P(X=r)=^nC_r p^rq^{n-r}

Where, p is the probability of success p=\frac{2}{7}

q is the probability of failure q=1-p, q=1-\frac{2}{7}=\frac{5}{7}

n is total number of trials n=8

r=3

Substitute the values,

P(X=3)=^8C_3 (\frac{2}{7})^3 (\frac{5}{7})^{8-3}

P(X=3)=\frac{8!}{3!5!}\times \frac{8}{343}\times (\frac{5}{7})^{5}

P(X=3)=\frac{8\times 7\times 6}{3\times 2\times 1}\times \frac{8}{343}\times \frac{3125}{16807}

P(X=3)=0.2428

Therefore, the probability that 8 mice are​ required is 0.2428.

5 0
3 years ago
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