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3 years ago

Find the height of a parallelogram with a base 6.75 meter and an area of 218.7 sqaure meters

Mathematics

2 answers:

Alex777 [14]3 years ago

7 0

Answer: 32.4

I hope this helps and have a wonderful day filled with joy!!

<3

maxonik [38]3 years ago

6 0

H= A/b
h=218.7/6.75
h=32.4

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Question 7 50 points

meriva

Answer:

1f 2×4 -3×³ 12k is divided by × - 3 the remainder is 21 find the value of k

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2 years ago

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There is 1/4 ounce of yeast in every 2 1/4 teaspoons of yeast. A recipe for bread calls for 2 teaspoons of yeast. How many ounce

tatyana61 [14]

Here is the set up:

(1/4)/2 = (9/4)/y

Solve for y.

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3 years ago

Consider the equation x2+4x+9=0 in standard form. Which equation shows the coefficients a, b, and c correctly substituted into t

Afina-wow [57]

Answer:

Step-by-step explanation:

The general form of a quadratic equation is ax²+bx+c = 0

Given the quadratic equation x²+4x+9=0 in its standard form, on comparing with the general equation we can get the value of the constant a, b and c as shown;

ax² = x²

a = 1

bx = 4x

b = 4

c = 9

The quadratic formula is given as x = -b±√(b²-4ac)/2a

Substituting the constant;

x = -4±√(4²-4(1)(9))/2(1)

x = -4 ±√(16-36)/2

x = -4±√-20/2

x = -4±(√-1*√20)/2

Note that √-1 = i

x = -4±(i√4*5)/2

x = (-4±i2√5)/2

x = -4/2±i2√5/2

x = -2±i√5

The solution to the quadratic equation are -2+i√5 and -2i-√5

5 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:

<em>F</em> = The word free occurs in an email

<em>S</em> = The email is spam

<em>V</em> = 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

12 yo<br> 5 yd<br> what is the length of the hypotenuse

Allushta [10]

I'm sure that the hypotenuse is 13 yards

Here's the formula used:

Length of Hypotenuse² = Length of side 1² + Length of side 2²

So here's how to solve this question:

12 squared + 5 squared = hypotenuse

(Then square the values)

144 + 25 = 169

Divide 169 by the square root to get the length of the hypotenuse: 13

Hope this helps!

7 0

3 years ago