Ask question

Ask question

All categories

2 years ago

11

Sole this addtion question by substituting numbers for the given letters, to find the value of START Each letter represents a un

ique digit.
JOAN
+ CAN
______

1 answer:

sladkih [1.3K]2 years ago

8 0

I’d say 69 i could be wrong tho :) or it is 12 hope this helps

You might be interested in

Can u please help me with 27

Pachacha [2.7K]

The answer would be 237.375. To get this answer, you will first use the formula given for each unit. First, you will do the first unit by putting 8 to the third power, or basically 8x8x8. After you get the product 512, you will then do the next unit. Do that one by doing 6.5 to the third power, or as followed, 6.5x6.5x6.5. Once you get the product 274.625, you will then subtract that from 512. You will then receive the answer 237.375.

5 0

3 years ago

30+ Points!!!<br> 5. Solve the following inequalities.<br> a) 2 log3x – 2 logx3 -3 <0

Ilya [14]

Answer:

I answered your last question also

2 log3x – 2 logx3 -3 <0

$\mathrm{Subtract\:}2\log ^3\left(x\right)\mathrm{\:from\:both\:sides}$

$2\log ^3\left(x\right)-2logx^3-3-2\log ^3\left(x\right)$

$\mathrm{Simplify}$

$-2logx^3-3$

$\mathrm{Add\:}3\mathrm{\:to\:both\:sides}$

$-2logx^3-3+3$

$\mathrm{Simplify}$

$-2logx^3$

$Multiply\:both\:sides\:by\:-1\:\left(reverse\:the\:inequality\right)$

$\left(-2logx^3\right)\left(-1\right)>-2\log ^3\left(x\right)\left(-1\right)+3\left(-1\right)$

$\mathrm{Simplify}$

$2lx^3og>2\log ^3\left(x\right)-3$

$\mathrm{Divide\:both\:sides\:by\:}2lx^3o;\quad \:l>0$

$\frac{2lx^3og}{2lx^3o}>\frac{2\log ^3\left(x\right)}{2lx^3o}-\frac{3}{2lx^3o};\quad \:l>0\\$

$\mathrm{Simplify}$

$g>\frac{2\log ^3\left(x\right)-3}{2lx^3o};\quad \:l>0$

Step-by-step explanation:

6 0

2 years ago

If you flip a coin and roll a six sided die what is the probability that you will flip a heads and roll at least a three

ruslelena [56]

Answer:

1/3

Step-by-step explanation:

H3,H4,H5,H6

p=1/2×1/6+1/2×1/6+1/2×1/6+1/2×1/6

=4×1/2×1/6=1/3

5 0

3 years ago

Which expression is equivalent to the area of metal sheet required to make this square-shaped traffic sign?

Lelechka [254]

Answer:

A is the correct answer

Step-by-step explanation:

Area= (8x-2)^2=(8x)^2 -2(8x)(2) +(2)^2

=64x-32x+4

7 0

2 years ago

The city has an average of 13 days of rainfall for April.

zhenek [66]

Using the Poisson distribution, we have that:

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

<h3>What is the Poisson distribution?</h3>

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

For this problem, the mean is given as follows:

$\mu = 13$

The probability of having exactly 10 days of precipitation in the month of April is P(X = 10), hence:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 10) = \frac{e^{-13}13^{10}}{(10)!} = 0.0859$

There is a 0.0859 = 8.59% probability of having exactly 10 days of precipitation in the month of April.

The probability of having less than three days of precipitation in the month of April is:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

In which:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-13}13^{0}}{(0)!} \ approx 0$

$P(X = 1) = \frac{e^{-13}13^{1}}{(1)!} = 0.00003$

$P(X = 2) = \frac{e^{-13}13^{2}}{(2)!} = 0.00019$

Then:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 0.00003 + 0.00019 = 0.00022

There is a 0.00022 = 0.022% probability of having less than three days of precipitation in the month of April.

For more than 15 days, the probability is:

P(X > 15) = P(X = 16) + P(X = 17) + ... + P(X = 20)

Applying the formula for each of these values and adding them, we have that P(X > 15) = 0.2364, hence:

There is a 0.2364 = 23.64% probability of having more than 15 days of precipitation in the month of April.

More can be learned about the Poisson distribution at brainly.com/question/13971530

#SPJ1

6 0

1 year ago