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3 years ago

Simplificar: 8 – 14 + 7 +9

Mathematics

2 answers:

lesantik [10]3 years ago

8 0

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jeka57 [31]3 years ago

6 0

Step-by-step explanation:

this is your answer.............

You might be interested in

Jenifer is 1 year older than twice Zack's age. Write an expression that repersents Jenifer's age in relation to zack.

Sedbober [7]

Answer:

j = 2z + 1

jennifer is one year older or jennifer is plus one.

twice zacks age or two times zacks age.

jennifer's age is two times zacks age plus one or

j = 2z + 1

3 0

3 years ago

What’s the correct answer for this?

erastovalidia [21]

Answer:

8/17

Step-by-step explanation:

7 0

3 years ago

Both the La Plata river dolphin (Pontoporia blainvillei) and

Citrus2011 [14]

Answer:

1. A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin; 2. The radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball; 3. The volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

Step-by-step explanation:

If we expressed a number as:

$\\ N = a * 10^{b}$ (1)

Where

$\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10}$ (2)

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The La Plata river dolphin weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the scientific notation):

$\\ 30kg \leq Dolphin_{weight} \leq 50kg$

$\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg$

$\\ 35000kg \leq Whale_{weight} \leq 40000kg$

$\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg$

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

$\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)$

$\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)$

Then

$\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}$

$\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}$

Thus

A sperm whale is 3 orders of magnitude heavier than a La Plata river dolphin.

<h3>Second case</h3>

Following the same reasoning, we can conclude that the radius of the larger ball is one (1) order of magnitude bigger than the radius of the smaller ball:

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}$

$\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}$

<h3>Third case</h3>

For this case, we need to calculate the volume of a sphere for both radii (1cm and 10cm).

The volume of a sphere is

$\\ V_{sphere} = \frac{4}{3}*\pi*R^{3}$

Then, the volume of the ball of radius 1cm is:

$\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}$

$\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}$

And, the volume of the ball of radius 10cm is:

$\\ V_{radius=10} = \frac{4}{3}*\pi*(10cm)^{3}$

$\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}$

Thus

$\\ \frac{10^{3}}{10^{0}} = 10^{3}$

As a result, the volume of the larger ball is 3 orders of magnitude bigger than the volume of the smaller ball.

4 0

3 years ago

If we wanted to create a new 90% confidence interval from a different sample for the proportion of those with a two on one date

slava [35]

Answer: 271

Step-by-step explanation:

The formula we use to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

, where $z_{\alpha/2}$ is the two-tailed z-value for significance level of $(\alpha)$

p = prior estimation of the proportion

E = Margin of error.

If prior estimation of the proportion is unknown, then we take p= 0.5 , the formula becomes

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2$

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2$

Given : Margin of error : E= 0.05

Confidence level = 90%

Significance level $\alpha=1-0.90=0.10$

Using z-value table , Two-tailed z-value for significance level of $0.10$

$z_{\alpha/2}=1.645$

Then, the required sample size would be :

$n=0.25(\dfrac{1.645}{0.05})^2$

Simplify,

$n=270.6025\approx271$

Hence, the required minimum sample size =271

3 0

3 years ago

ANSWER ASAPPP PLSSS. ILL MARK BRAINLIEST TOO

nevsk [136]

Answer:

Step-by-step explanation:

11.04 = 10(1.02)^n

1.104 = 1.02^n

ln 1.104 = ln 1.02^n

ln 1.104 = n ln 1.02

n = ln 1.104/ ln 1.02

n = 4.99630409516

4.99 can be rounded to 5.

So a reasonable domain would be 0 ≤ x < 5

PART B)

f(0) = 10(1.02)^0

f(0) = 10(1)

f(0) = 10

The y-intercept represents the height of the plant when they began the experiment.

f(1) = 10(1.02)^1

f(1) = 10(1.02)

f(1) = 10.2

(1, 10.2)

f(5) = 10(1.02)^5

f(5) = 10(1.1040808)

f(5) = 11.040808

f(1)=10(1.02)^1

f(1)=10.2

Average rate= (fn2-fn1)/(n2-n1)

=11.04-10.2/(5-1)

=0.22

the average rate of change of the function f(n) from n = 1 to n = 5 is 0.22.

3 0

2 years ago

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Simplificar: 8 – 14 + 7 +9 ​

Simplificar: 8 – 14 + 7 +9