All categories

3 years ago

Please help will mark brainliest

Mathematics

1 answer:

Alja [10]3 years ago

3 0

Answer:

17 cm²

Step-by-step explanation:

First I cut this figure into 2 rectangles. One big and small one. (the big one is on top of the small rectangle).

For the big rectangle, I multiplied its length by its width.

The area of the big rectangle: 3cm * 5cm = 15cm²

For the smaller rectangle, I also multiplied its length and width.

The area of the small rectangle: 1cm * 2 cm = 2cm²

Add up its areas: 15+2

To get 17cm² as your final answer

You might be interested in

There are 250 bricks used to build a wall 20 feet high.How many bricks will be used to buld a wall that is 30 fet high

Dmitry_Shevchenko [17]

Answer: 375

Step-by-step explanation:

Since we are given the information that there are 250 bricks used to build a wall 20 feet high, the number of brick used per feet will be:

= 250/20

= 12.5 bricks per feet.

Therefore, the number of bricks that will be used to buld a wall that is 30 fet high will be:

= 12.5 × 30

= 375 bricks.

Therefore, 375 bricks would be used.

3 0

3 years ago

choose an equation in slope-intercept form for the line that passes through (-8,1) and is perpendicular to the y=2x-17. select o

Fed [463]

Y = 2x - 17, comparing to y = mx + c, slope m = 2.

If perpendicular, the new slope would be -1/2, that is the negative reciprocal of 2.

And passing through (-8 , 1).

using y = mx + c, and x = -8, y = 1, m = -1/2

1 = -1/2*-8 + c

1 = 4 + c

1 - 4 = c

c = -3

y = mx + c, substituting m = -1/2, and c = -3, y = -(1/2)x - 3.

Option C.

5 0

3 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.