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2 years ago

Simplify (+8)-(-4)+(-5)

Mathematics

2 answers:

Mumz [18]2 years ago

4 0

Answer:

Step-by-step explanation:

(+8)-(-4)+(-5)

8+4-5

12-5

melisa1 [442]2 years ago

4 0

Answer:

Step-by-step explanation:

8+4+5=17

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At Marco's school, 5/8 of the students are in the band. What percent of the students are in the band?

Harman [31]

62.5%

5/8 = 0.625 then multiply by 100% to get in terms of percent of 62.5%

7 0

2 years ago

A pill has the shape of a cylinder with a hemisphere at each end. The height of the cylindrical portion is 12mm and the overall

Volgvan

Answer:

The volume of the pill is $V=452\ mm^{3}$

Step-by-step explanation:

Find the volume of the pill in cubic millimeters

we know that

The volume of the pill is equal to the volume of the cylinder plus the volume of a sphere (two hemisphere is equal to one sphere)

we have

$V=\frac{4}{3}\pi r^{3}+\pi r^{2}h$

$r= (18-12)/2=3\ mm\\ h=12\ mm$

assume

$\pi =3.14$

substitute

$V=\frac{4}{3}(3.14)(3)^{3}+(3.14)(3)^{2}(12)\\V=452\ mm^{3}$

8 0

3 years ago

Pls welp will give brainliest to whoever answers correctly first

Sholpan [36]

There are 12 squares in each 4 x 3 grid.

25% of 12 = 3

This means that there should be 3 shaded squares in the grid.

Your answer is B. because it has 3 shaded squares.

7 0

2 years ago

Read 2 more answers

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

A shoe manufacturer was investigating the weights of men's soccer cleats. He felt that the weight of these cleats was less than

aleksklad [387]

Answer:

The conclusion is that the researcher was correct

Step-by-step explanation:

From the question we are told that

The sample size is $n = 13$

The sample mean is $\= x = 9.63$

The standard deviation is $s = 0.585$

The significance level is $\alpha = 0.05$

The Null Hypothesis is $H_o : \mu = 0$

The Alternative Hypothesis is $H_a = \mu < 10$

The test statistic is mathematically represented as

$t = \frac{\= x - \mu }{\frac{s}{\sqrt{n} } }$

Substituting values

$t = \frac{9.63 - 10 }{\frac{0.585}{\sqrt{13} } }$

$t = - 2.280$

Now the critical value for $\alpha$ is

$t_{\alpha } = 1.645$

This obtained from the critical value table

So comparing the critical value of alpha and the test value we see that the test value is less than the critical value so the Null Hypothesis is rejected

The conclusion is that the researcher was correct

6 0

3 years ago

Simplify (+8)-(-4)+(-5)​

Simplify (+8)-(-4)+(-5)