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3 years ago

15/16 as a decimal rounded to the nearest tenth

Mathematics

1 answer:

matrenka [14]3 years ago

6 0

Answer:

0.9

Step-by-step explanation:

$\frac{15}{16}=0.9375\approx0.9$

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Suppose a random sample of n measurements is selected from a binomial population with probability of success p = .39. Given n =

Oksi-84 [34.3K]

Answer:

The answer is below

Step-by-step explanation:

a) Given that the number of sample (n) = 300, therefore since n > 30, the distribution of the sample means is going to be normally distributed.

b) The mean of the distribution of sample means (also known as the Expected value of M) is equal to the population mean μ.

$\mu_x=\mu=np=300*0.39=117$

c) The standard deviation of the distribution of sample means is called the Standard Error of M, it is given by:

$\sigma_x=\sqrt{np(1-p)} =\sqrt{300*0.39(1-0.39)} =8.44$

8 0

3 years ago

Molly had a $12 gift card to download music. She bought 8 songs that each cost the same amount. Write a variable equation, then

inn [45]

Answer:

1.Equation ----- 8x=12

2. x=1.5

3. cost of a single song is $1.5

4.$12 gift card will download 8 songs

Step-by-step explanation:

The amount of money the gift card holds is $12

Number of songs bought= 8 songs

Let the cost of a single song to be =$x

The cost of 8 songs will be =$x × 8=$ 8x

So the 8 songs will all cost $12

This means; $ 8x = $12

Equation------ 8x=12

Finding cost of a single song

8x=12------------------divide by 8 both sides

x=12/8------------------simplify

x=3/2 = $ 1.5

Checking for solution

if a single song costs $1.5 , who many songs can a gift card with $12 download?

Here we divide $12 by $1.5 = $12/$1.5= 8 songs------------correct number of songs as mentioned in the question.

7 0

3 years ago

How do you solve the equation F = 9/5(K - 273.15) + 32 when you're solving for K?

Mumz [18]

$\displaystyle\\Answer:\ K=\frac{5}{9}(F-32^0)+273.15^0$

Step-by-step explanation:

$\displaystyle\\F=\frac{9}{5} (K-273.15)+32\\F-32=\frac{9}{5} (K-273.15)+32-32\\F-32=\frac{9}{5} (k-273.15)\\Multiply\ both\ parts\ of\ the\ equation\ by\ \frac{5}{9} \\(\frac{5}{9})(F-32)=\frac{9}{5} (K-273.15)(\frac{5}{9} )\\\\\frac{5}{9} (F-32)=K-273.15\\\\\frac{5}{9} (F-32)+273.15=K-273.15+273.15\\\\\frac{5}{9} (F-32)+273.15=K$

7 0

2 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago

The length of a rectangle is 5 in. more than 3 times its width. the area of the rectangle is 50in.2. a quadratic function repres

Goshia [24]

The domain is actually the x value of the function, so we need to find the value of x
suppose the width is x, the length is then 3x+5
the area is 50^2 inches, so x(3x+5)=50 => 3x^2+5x-50=0
Factor this quadratic equation: (x-5)(3x+10)=0 =>x=5 or x=-10/3
width can not be negative, so the width is 5
the domain is x=5

6 0

3 years ago