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3 years ago

60÷360×3.14×15×15-√3÷4×15×15

Mathematics

1 answer:

alina1380 [7]3 years ago

4 0

Step-by-step explanation:

The answer to this question will be 20.322142074...

hope it helps

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A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a s

Umnica [9.8K]

Answer:

88.51 is the minimum score needed to receive a grade of A.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 73

Standard Deviation, σ = 11

We are given that the distribution of exam grades is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.0793.

$P( X \geq x) = P( z \geq \displaystyle\frac{x - 73}{11})=0.0793$

$= 1 -P( z \leq \displaystyle\frac{x - 73}{11})=0.0793$

$=P( z \leq \displaystyle\frac{x - 73}{11})=0.9207$

Calculation the value from standard normal z table, we have,

$P(z \leq 1.410) = 0.9207$

$\displaystyle\frac{x - 73}{11} = 1.410\\x =88.51$

Hence, 88.51 is the minimum score needed to receive a grade of A.

3 0

2 years ago

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

2 years ago

The Midpoint of AB dkdldl

Nutka1998 [239]

The Midpoint of AB is (1,0).

Given that:

In line AB, where the coordinates of A is (3,1) and coordinates of B is (-1,-1).

To find:

The Midpoint of line AB.

So, according the question

We know that,

The midpoint M of a line segment AB with endpoints A (x₁, y₁) and B (x₂, y₂) has the coordinates M ( $\frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2}$ ).

Now from question,

We know that the the coordinates of A is (3,1) and coordinates of B is (-1,-1) of line AB.

So, we can say that

A is (3,1) or x₁ = 3 and y₁ = 1.

B (-1,-1) or x₂ = -1 and y₂ = -1.

∵ The coordinates of midpoint M (X,Y)

X = $\frac{x_1 + x_2}{2}$

= $\frac{3-1}{2}$

= 2/2

X = 1.

And

Y = $\frac{y_1 + y_2}{2}$

= $\frac{1-1}{2}$

= 0/2

Y = 0.

So, the midpoint of line AB is M (1,0)

To learn more about Midpoint of line, please click on the link;

brainly.com/question/14687140

#SPJ1

8 0

1 year ago

Ryan fills a bath with

S_A_V [24]

Answer:

38.8 litres

Step-by-step explanation:

22.6 + 18.7 = 41.3

41.3 - 2.5 = 38.8

38.8 litres

4 0

1 year ago

Read 2 more answers

Can you graph a line if its slope is undefined? Explain.

JulijaS [17]

You can but it would be going straight up and down

6 0

3 years ago

60÷360×3.14×15×15-√3÷4×15×15​

60÷360×3.14×15×15-√3÷4×15×15