All categories

3 years ago

Find the product of 0.056 and 0.73. 0.04088 0.4088 0.0056

Mathematics

2 answers:

alekssr [168]3 years ago

3 0

we know that

$0.056= 56*10^{-3}\\ 0.73=73* *10^{-2}$

$0.056*0.73=56*10^{-3}*73*10^{-2}$

$=(10^{-5})*(56*73)$

$=(10^{-5})*[(50+6)*(70+3)]$

$=(10^{-5})*[(3,500+150+420+18)]$

$=(10^{-5})*[(3,500+588)]$

$=(10^{-5})*[(4,088)]$

$=0.04088$

therefore

<u>the answer is</u>

$=0.04088$

Verizon [17]3 years ago

3 0

0.056 * 0.73 = 0.04088

You might be interested in

Given that f(x)=(k x)/(2x 1)and that f(-3)=2, find the value of k

makvit [3.9K]

Answer:

Kindly upload a better version

Step-by-step explanation:

3 0

3 years ago

(4.1.4) Let X and Y be Bernoulli random variables. Let Z = X + Y. a. Show that if X and Y cannot both be equal to 1, then Z is a

Fynjy0 [20]

Step-by-step explanation:

Given that,

X ~ Bernoulli $(p_x)$ and Y ~ Bernoulli $(y_x)$

X + Y = Z

The possible value for Z are Z = 0 when X = 0 and Y = 0

and Z = 1 when X = 0 and Y = 1 or when X = 1 and Y = 0

If X and Y can not be both equal to 1 , then the probability mass function of the random variable Z takes on the value of 0 for any value of Z other than 0 and 1,

Therefore Z is a Bernoulli random variable

If X and Y can not be both equal to 1

then,

$p_z = P(X=1$ or $Y=1)\\$

$p_z = P(X=1)+P(Y=1)-P(=1$ and $Y =1)$

$p_z = P(x=1)+P(Y=1)\\\\p_z=p_x+p_y$

If both X = 1 and Y = 1 then Z = 2

The possible values of the random variable Z are 0, 1 and 2.

since a Bernoulli variable should be take on only values 0 and 1 the random variable Z does not have Bernoulli distribution

7 0

3 years ago

(a^-3bc^0)^-2/(ab^3c^4)^5

mr Goodwill [35]

6b6
——— (hope this helps)
ac2

5 0

3 years ago

Write a ratio, in the simplest form and a percent for the shaded area (there's a 5x4 box and 6 cubes are shaded)

Alexxandr [17]

Answer:

$\%Percent = 30 \%$

Step-by-step explanation:

Given

$Shaded = 6\ cubes$

$Box = 5\ x\ 4$

Required

Determine the percentage of the shaded cubes

First, we calculate the number of boxes:

$Box = 5\ x\ 4$

Express as multiplication

$Box = 5\ *\ 4$

$Box = 20$

This implies that there are 20 cubes in the box.

The percentage of the shaded cubes is:

$\%Percent = \frac{Shaded}{Box} * 100\%$

$\%Percent = \frac{6}{20} * 100\%$

$\%Percent = \frac{6* 100}{20} \%$

$\%Percent = \frac{600}{20} \%$

$\%Percent = 30 \%$

<em>30% were shaded</em>

5 0

3 years ago

The mass of a species of mouse commonly found in houses is normally distributed with a mean of 20.2 grams with a standard deviat

Sophie [7]

Answer:

12.1%

Step-by-step explanation:

Given that:

Mean (μ) = 20.2 grams and standard deviation (σ) = 0.18 grams.

The z score is a score used to determine the number of standard deviations by which the raw score is above or below the mean. A positive z score means that the raw score is above the mean and a negative z score means that the raw score is below the mean. It is given by:

$z=\frac{x-\mu}{\sigma}$

a) For x < 19.99 g:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{19.99-20.2}{0.18} \\\\z=-1.17$

From the normal distribution table, P(x < 19.99) = P(z < -1.17) = 0.1210 = 12.1%

The probability that a randomly chosen mouse has a mass of less than 19.99 grams is 12.1%

6 0

3 years ago