All categories

3 years ago

A photograph of sides 35cm by 22cm is mounted onto a

Mathematics

2 answers:

trapecia [35]3 years ago

7 0

<h3>Answer: 580 square cm</h3>

===================================================

Work Shown:

A = area of smaller rectangle

A = length*width

A = 35*22

A = 770

B = area of larger rectangle

B = length*width

B = 45*30

B = 1350

C = area of the border only

C = B - A

C = 1350-770

C = 580 square cm

KengaRu [80]3 years ago

3 0

Step-by-step explanation:

Dimension of photograph is 35cm and 22cm.

And external dimension of photo frame is 45cm and 30cm

So, the area of the border surrounding the photograph=Area of photo frame−Area of photo.

So, The area of the border surrounding the photograph=45×30−35×22

=1350−770=580cm

You might be interested in

Please help I'm very confused and frustrated.

Ray Of Light [21]

Step-by-step explanation:

-36.8 + 9.2(2). [as the value of x is 2]

-36.8 + 18.4

= - 18.4 [as the number with the

Negative sign is greater]

7 0

3 years ago

Read 2 more answers

Use synthetic division to find the quotient of (44 + 22 − 4 + 12) ÷ ( + 2). Show your work for full credit.

valina [46]

Just remember BODMAS, which B= brackets O= Order D= Division M= Multiply A=Addition and S=Subtract

So first you solve the brackets so 44+22 is 66

So 66-4+12 so 66-4 is 62

62+12 is 74

So now you have got 74 divided by 2 which is 37

Hope this helps xx

5 0

3 years ago

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago

What is the answer to −9−(−13) its integers

Setler [38]

Answer: 4

Step-by-step explanation:

6 0

3 years ago

Write the linear equation that gives the rule for this table.

Maksim231197 [3]

to get the equation of any straight line all we need is two points, well, let's just grab them from the table.

hmmm let's use (-49 , -39) and hmmm say (67 , 77)

$(\stackrel{x_1}{-49}~,~\stackrel{y_1}{-39})\qquad (\stackrel{x_2}{67}~,~\stackrel{y_2}{77}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{77}-\stackrel{y1}{(-39)}}}{\underset{run} {\underset{x_2}{67}-\underset{x_1}{(-49)}}}\implies \cfrac{77+39}{67+49}\implies \cfrac{116}{116}\implies 1$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-39)}=\stackrel{m}{1}(x-\stackrel{x_1}{(-49)}) \\\\\\ y+39=1(x+49)\implies y+39=x+49\implies y=x+10$

8 0

2 years ago