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3 years ago

PLEASE HELP (Giving Good Points!)

Mathematics

1 answer:

Vinvika [58]3 years ago

7 0

Answer: q=16

Step-by-step explanation:

if we suppose q=16

(q+4)/2=10

$(16+4)/2=10\\20/2=10\\10=10$

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There are 20 deer per 2 square miles In

sladkih [1.3K]

Answer:

594,250 (a)

Step-by-step explanation:

2 sq. miles = 20 deers

1 sq. mile = 20÷2 = 10

59,425 sq. miles = 59,425×10

= 594,250

8 0

3 years ago

HELP HELP HELP

mojhsa [17]

Answer:

2998 ; 17%

Step-by-step explanation:

Given the function:

t(c)=-3970.9(ln c)

c = % of carbon remaining ; t = time

1.) c = 47% = 47/100 = 0.47

t(0.47) = - 3970.9(In 0.47)

t = - 3970.9 * −0.755022

t = 2998.119

t = 2998

B.)

t = 7000

t(c)=-3970.9(ln c)

7000 = - 3970.9(In c)

7000 / - 3970.9 = In c

−1.762824 = In c

c = exp(−1.762824)

c = 0.1715596

c = 0.1715596 * 100%

c = 17.156% ; c = 17%

8 0

3 years ago

Read 2 more answers

What is the ordinate of the point (-9,4)?

statuscvo [17]

Answer:

-9

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

Rewrite each of the following statements in the form "∀ _____ x, _____." (a) All dinosaurs are extinct. ∀ x, . (b) Every real nu

jasenka [17]

Answer:

See below

Step-by-step explanation:

Essentially, we have to replace "quantifier words" like "All", "Every" by the universal quantifier ∀.

a) ∀ dinosaur x, x is extinct.

b) ∀ real number x, x is positive, negative, or zero.

c) ∀ irrational number x, x is not an integer.

d) ∀ logician x, x is not lazy.

e) ∀ integer x, x²≠ 2,147,581,953.

f) ∀ real number x, x²≠ -1.

In a) and b) we replace the words without major changes. In the other statements, we modify the statement using negation. For example, "No irrational numbers are integers." is equivalent to "Every irrational number is not integer".

8 0

3 years ago

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

4 years ago