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4 years ago

What is the largest sum your can get when you add 2 3 digit numbers

Mathematics

2 answers:

Ksju [112]4 years ago

8 0

The biggest 3 digits whole number is 999.

Hence, the biggest sum one can do with two of those is 999+999=1998

Serhud [2]4 years ago

3 0

1998 (999 + 999) because to get a really large number with two three digit numbers, you would add the two greatest three digit numbers.

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A sequence is defined recursively using the formula f(n + 1) = –0.5 f(n) . If the first term of the sequence is 120, what is f(5

Georgia [21]

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A soda manufacturer knows that sales of soda increase during the summer. To make sure that they get a large portion of the sales

Alexxx [7]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is

X: Number of winning caps out of 5 bottles.

Checking the binomial criteria:

- Number of trials is fixed n=5 bottles of soda

- Only two possible outcomes per trial: "success": winning cap; "failure": regular cap.

- The trial are independent.

- The probability of success is constant trough the experiment p= 1/12= 0.083

This variable has a binomial distribution with parameters n= 5 and p= 0.083

You have to calculate the probability of getting 4 out of 5 winning caps

P(X=4)

Using the following formula:

$P(X=4)= \frac{n!}{(n-X)!X!} * (p)^{X} * (q)^{n-X}= \frac{5!}{1!4!} * (0.083)^{4} * (0.917)^{1}= 0.0002$

Or using the tables for accumulated probabilities:

P(X=4)= P(X≤4)-P(X≤3)= 0.9999 - 0.9997= 0.0002

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3 years ago

g Which of the following is true about a p-value? Group of answer choices It measures the probability that the null hypothesis i

scZoUnD [109]

Answer:

It measures the probability of observing your test statistic, assuming the null hypothesis is true.

Step-by-step explanation:

The p-value, also known as the probability value <u>measures the probability of observing your test statistic, assuming the null hypothesis is true.</u>

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3 years ago

94b + 48 < b + 54 < 94b + 82

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DATES: The rule amendments will become effective October 29, 1999. Section IV of this release contains information on compliance dates.

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3 years ago

A graphing calculator is recommended. A crystal growth furnace is used in research to determine how best to manufacture crystals

Sophie [7]

Answer:

a) $w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

b) $202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

c)For this case $\epsilon = \pm 1$ since that's the tolerance 1C

d) $\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smalles value on this case $\delta =0.113$

e) For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

Step-by-step explanation:

For this case we have the following function

$T(w)= 0.1 w^2 +2.156 w +20$

Where T represent the temperature in Celsius and w the power input in watts.

Part a

For this case we need to find the value of w that makes the temperature 203C, so we can set the following equation:

$203= 0.1w^2 +2.156 w +20$

And we can rewrite the expression like this:

$0.1w^2 +2.156 w-183=$

And we can solve this using the quadratic formula given by:

$w =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

Where a =0.1, b =2.156 and c=-183. If we replace we got:

$w_1 = \frac{-2.156 -\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=-54.896$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-183)}}{2*0.1}=33.336$

And since the power can't be negative then the solution would be w = 33.34 watts.

Part b

For this case we can find the values of w for the temperatures 203-1= 202C and 203+1 = 204 C. And we got this:

$202= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-182=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-182)}}{2*0.1}=33.222$

$204= 0.1w^2 +2.156 w +20$

$0.1w^2 +2.156 w-184=$

$w_2 = \frac{-2.156 +\sqrt{2.156^2 -4(0.1)(-184)}}{2*0.1}=33.449$

So then the range of voltage would be between 33.22 W and 33.45 W.

Part c

For this case $\epsilon = \pm 1$ since that's the tolerance 1C

Part d

For this case we can do this:

$\delta_1 =|33.222-33.336|=0.116$

$\delta_2 =|33.449-33.336|=0.113$

So then we select the smallest value on this case $\delta =0.113$

Part e

For this case if we assume a tolerance of $\epsilon=\pm 1C$ for the temperature and a tolerance for the power input $\delta =0.113$ we see that:

$lim_{x \to a} f(x) =L$

Where a = 33.34 W, f(x) represent the temperature, x represent the input power and L = 203C

8 0

4 years ago