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2 years ago

Plz I need the answer very fast

Mathematics

1 answer:

baherus [9]2 years ago

3 0

Answer:

1. 5√32)^3

2^3=6

2. 5^2(1.5)

5^3=125

3. (0.00001)^2

0.00001×0.00001=0.0000000001

4. 3√343)^2

7^2=49

5. 3√216)^4

6^4=1296

6.19^0=1

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What is the measure of mop

marysya [2.9K]

61 degrees, hope this helps!

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3 years ago

The forth term of an arithmetic seaquence is 19 and the fifth term is 13 . what is the third term

Vaselesa [24]

The third term was likely to be 25. You can see a pattern -6 when it goes up a term.

5 0

2 years ago

Find the sum of the constants a, h, and k such that

ra1l [238]

Answer:

Step-by-step explanation:

The value of "a" is the coefficient of x^2, so we know that is 2.

Solve for h

Now, we have ...

2x^2 -8x +7 = 2(x -h)^2 +k

Expanding the right side gives us ...

= 2(x^2 -2hx +h^2) +k

= 2x^2 -4hx +2h^2 +k

Comparing x-terms, we see ...

-4hx = -8x

h = (-8x)/(-4x) = 2

Solve for k

Now, we're left with ...

2h^2 +k = 7 = 2(2^2) +k = 8 +k

Subtracting 8 we find k to be ...

k = 7 -8 = -1

And the sum of constants a, h, and k is ...

a +h +k = 2 +2 -1 = 3

The sum of the constants is 3.

6 0

3 years ago

An e-mail filter is planned to separate valid e-mails from spam. The word free occurs in 60% of the spam messages and only 4% of

ANEK [815]

Answer:

(a) 0.152

(b) 0.789

Step-by-step explanation:

Let's denote the events as follows:

F = The word free occurs in an email

S = The email is spam

V = The email is valid.

The information provided to us are:

The probability of the word free occurring in a spam message is, $P(F|S)=0.60$
The probability of the word free occurring in a valid message is, $P(F|V)=0.04$
The probability of spam messages is,

$P(S)=0.20$

First let's compute the probability of valid messages:

$P (V) = 1 - P(S)\\=1-0.20\\=0.80$

(a)

To compute the probability of messages that contains the word free use the rule of total probability.

The rule of total probability is:

$P(A)=P(A|B)P(B)+P(A|B^{c})P(B^{c})$

The probability that a message contains the word free is:

$P(F)=P(F|S)P(S)+P(F|V)P(V)\\=(0.60*0.20)+(0.04*0.80)\\=0.152\\$

The probability of a message containing the word free is 0.152

(b)

To compute the probability of messages that are spam given that they contain the word free use the Bayes' Theorem.

The Bayes' theorem is used to determine the probability of an event based on the fact that another event has already occurred. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is spam provided that it contains free is:

$P(S|F)=\frac{P(F|S)P(S)}{P(F)}\\=\frac{0.60*0.20}{0.152} \\=0.78947\\$

The probability that a message is spam provided that it contains free is approximately 0.789.

(c)

To compute the probability of messages that are valid given that they do not contain the word free use the Bayes' Theorem. That is,

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

The probability that a message is valid provided that it does not contain free is:

$P(V|F^{c})=\frac{P(F^{c}|V)P(V)}{P(F^{c})} \\=\frac{(1-P(F|V))P(V)}{1-P(F)}\\=\frac{(1-0.04)*0.80}{1-0.152} \\=0.90566$

The probability that a message is valid provided that it does not contain free is approximately 0.906.

4 0

3 years ago

Pls help I won't be able to watch you tube this summer. Select all the common denominators for 1/4 and 7/10. Explain.

stepan [7]

The common denominators of ¹/₄ and ⁷/₁₀ are; 20 and 40

<h3>How to find the denominator?</h3>

The numerator of a fraction is the top value while the denominator of a fraction is the value at the bottom.

Now, we want to find the common denominator of ¹/₄ and ⁷/₁₀.

The common denominator can't be less than 4 or 10 and so has to be at least two terms that are divisible by 4 and 10. The next two common denominators that are greater than 10 but divisible by both 4 and 10 are 20.

Read more about denominators at; brainly.com/question/12182689

#SPJ1

5 0

1 year ago

Plz I need the answer very fast​

Plz I need the answer very fast