Substitute: StartFraction sine (45 degrees) Over a EndFraction = StartFraction sine (77 degrees) Over 8 EndFraction .

Mathematics

1 answer:

fiasKO [112]3 years ago

6 0

Answer:

a=5.81

Step-by-step explanation:

We want to solve for the value of a in the expression:

$\frac{sin 45}{a}=\frac{sin 77}{8}$

Cross Multiply

8 X sin 45=a X sin 77

Divide both sides by sin 77° to obtain a.

$a=\frac{8 X sin 45}{sin 77}$

a=5.81

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oscar bought some markers notebooks and packs of sticky notes for his office he bought the same number of markers as notebooks h

s2008m [1.1K]

The number of each kind of office supply oscar bought is = markers= (43.2 - 4.2y)/3 notebooks = (43.2-3x)/4.2, sticky notes = [43.2 - 4.2y)/3 + (43.2-3x)/4.2] + 3.

<h3>Calculation of total office supply</h3>

The number of markers he bought = X

The number of notebooks he bought= y

The number of sticky notes he bought= z= 3 + X+y

The amount he paid for markers = $1.05*X

The amount he paid for notebooks = $2.25*y

The amount he paid for sticky notes= $1.95 (3 + X+y)

The total amount he paid is = $49.05

That is;

$49.05 = $1.05x + $2.25y + $1.95 (3 + X+y)

$49.05 = $1.05x + $2.25y + $5.85 + $1.95x + $1.95y

$49.05 - $5.85 = 3x + 4.2y

$43.2 = 3x + 4.2y

Make X the subject of formula,

3x = 43.2 - 4.2y

X = (43.2 - 4.2y)/3

y= (43.2-3x)/4.2

z = [43.2 - 4.2y)/3 + (43.2-3x)/4.2] + 3

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1 year ago

The lengths of a particular snake are approximately normally distributed with a given mean mu = 15 in. and standard deviation si

Helga [31]

The percentage of the snakes is longer than 16.6 in. with a mean of 15 in. and a standard deviation of 0.8 in. is 2.275%.

<h3>What is a normal distribution?</h3>

It is also called the Gaussian Distribution. It is the most important continuous probability distribution. The curve looks like a bell, so it is also called a bell curve.

The z-score is a numerical measurement used in statistics of the value's relationship to the mean of a group of values, measured in terms of standards from the mean.

The lengths of a particular snake are approximately normally distributed with a given mean = 15 in. and standard deviation = 0.8 in.

Then the percentage of the snakes is longer than 16.6 in. will be

$z = \dfrac{ x - \mu }{ \sigma }\\\\z = \dfrac{16.6-15}{0.8}\\\\z = 2$