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3 years ago

Whats the sum of (-3)+(-5)

Mathematics

2 answers:

Flura [38]3 years ago

8 0

the answer is -8. have a nice day!⭐️

mrs_skeptik [129]3 years ago

6 0

Answer:

-8

Step-by-step explanation:

If you are adding two negative numbers its basically just normal addition but you add the negative sign before. So how i do it is 5+3=8 but with invisible negative signs. So really its -8

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Repeated student samples. Of all freshman at a large college, 16% made the dean’s list in the current year. As part of a class p

Dima020 [189]

Answer:

a) p-hat (sampling distribution of sample proportions)

b) Symmetric

c) σ=0.058

d) Standard error

e) If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

Step-by-step explanation:

a) This distribution is called the <em>sampling distribution of sample proportions</em> <em>(p-hat)</em>.

b) The shape of this distribution is expected to somewhat normal, symmetrical and centered around 16%.

This happens because the expected sample proportion is 0.16. Some samples will have a proportion over 0.16 and others below, but the most of them will be around the population mean. In other words, the sample proportions is a non-biased estimator of the population proportion.

c) The variability of this distribution, represented by the standard error, is:

$\sigma=\sqrt{p(1-p)/n}=\sqrt{0.16*0.84/40}=0.058$

d) The formal name is Standard error.

e) If we divided the variability of the distribution with sample size n=90 to the variability of the distribution with sample size n=40, we have:

$\frac{\sigma_{90}}{\sigma_{40}}=\frac{\sqrt{p(1-p)/n_{90}} }{\sqrt{p(1-p)/n_{40}}}}= \sqrt{\frac{1/n_{90}}{1/n_{40}}}=\sqrt{\frac{1/90}{1/40}}=\sqrt{0.444}= 0.667$

If we increase the sample size from 40 to 90 students, the standard error becomes two thirds of the previous standard error (se=0.667).

0 0

3 years ago

Which applies the power of a power rule to simplify the expression (8 Superscript 12 Baseline) cubed?

rodikova [14]

Answer: (8^{12})^3=8^{12\times 3}=8^{36}

Step-by-step explanation:

Given : the expression (8^{12})^3

We have to simplify the given expression and choose the correct from the given options.

Consider the expression (8^{12})^3

Using property of exponents,

\left(a^b\right)^c=a^{b\times c}

We have,

(8^{12})^3=8^{12\times 3}=8^{36}

7 1

4 years ago

Read 2 more answers

Help please 23 points will make brainliest

svlad2 [7]

It should be 78.54.
Double check that.
Formula: V=πr2h/3
r2 is R squared

8 0

3 years ago

You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equal mass (3,000 kg each)

Kazeer [188]

Answer:

1) Kinetic energy of the truck 1 is 600,000 joules.

Kinetic energy of the truck 2 is 1,837,500 joules.

2 a)Potential energy of truck 1 is 646,800 Joules.

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

Step-by-step explanation:

Mass of the truck 1 and truck 2,m = 3000 kg

Velocity of the truck 1, $v_1$ = = 20 m./s

Velocity of the truck 2, $v_2$ = 35 m./s

1. Kinetic energy of the truck 1: $\frac{1}{2}mv_{1}^2=\frac{1}{2}\times 3000 kg\times (20 m/s)^=600,000 joules$

Kinetic energy of the truck 2: $\frac{1}{2}mv_{2}^2=\frac{1}{2}\times 3000 kg\times (35 m/s)^=1,837,500 joules$

2.a) Energy loose by the truck 1:

Initial height of the truck on the hill = 22 m

At that height potential energy will be maximum:

Potential energy of truck 1 at that height = $mgh=3000\times 9.8 m/s^2\times 22 m=646,800 Joules$

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

3 0

4 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

7 0

3 years ago

Whats the sum of (-3)+(-5)​

Whats the sum of (-3)+(-5)