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Nady [450]
3 years ago
13

The rectangular top of a small box has dimensions of 140 mm by 230 mm. How long (in centimeters) would a ribbon need to be to go

all the way around the edge of the top of the box?
Mathematics
2 answers:
egoroff_w [7]3 years ago
8 0

Answer:

74

Step-by-step explanation:

i got 100 on test

Bas_tet [7]3 years ago
6 0

Answer:

74 cm would a ribbon need to be to go all the way around the edge of the top of the box .

Step-by-step explanation:

Formula

Perimeter of a rectangle = 2 (Length + Breadth)

As given

The rectangular top of a small box has dimensions of 140 mm by 230 mm.

Perimeter of the rectangle box top = 2 × (140 + 230)

                                                          = 2 × 370

                                                          = 740 mm

As

1 mm = 0.1 cm

Now change into 740 mm into cm.

740 mm = 0.1 × 740  

              = 74 cm

Perimeter of the rectangle box top = 74 cm

Therefore 74 cm would a ribbon need to be to go all the way around the edge of the top of the box .

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In a city with a population of 1 million, 500 people have been diagnosed with cancer, whereas the rest of the people do not have
Leni [432]

Class distribution is considered to be normal using normal approximation to binomial.

A frequency distribution is a table that shows classes or intervals of data with a count of the number in each class.

Given :

In a city with a population of 1 million

i.e Population size (N) = 1 million = 1000000

From which 500 people have been diagnosed with cancer

Let X : No. of people  have been diagnosed with cancer

X = 500

Population proportion (p) :

p = XNp =5001000000p = 0.0005

So , the proportion of people who do not have cancer (q)=1-p

                                                                                               =1-0.0005

                                                                                               =0.9995

Here the distribution of X is Binomial

Such that :

i. No. of trials are fixed

ii. Probability of success is constant (p=0.0005)

iii. Possible outcomes are two (having cancer or not)

Here we can use normal approximation to Binomial

Because N is very large

So  class distribution is considered to be normal using normal approximation to binomial.

To learn more, visit:

brainly.com/question/9325204

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5 0
1 year ago
What is the equation of the line in y=mx+b format? BRAINLIST WILL BE REWARDED
harkovskaia [24]

Answer:

y = 2/5x - 1

Step-by-step explanation:

There are a couple of different ways to figure this out, but since you have a graph, the fastest way is to count the slope, or m in the formula, by counting the rise (units up) over run (units to the right), which is 2/5. Then, find the value of b by finding the y-intercept on the graph, or -1.

3 0
3 years ago
Plot (4, −34) on the coordinate plane.
Lyrx [107]

Answer:

None

Step-by-step explanation:

Think about it. 4 is the x axis. -34 is the y axis. It will be located in quadrant IV.

8 0
3 years ago
Read 2 more answers
please help. answer is 62.1m(squared) and the objects are composite! please show the work, no matter what i do i cant seem to ge
yuradex [85]

Answer:

SA=62.1\ m^2

Step-by-step explanation:

step 1

Find the surface area of the smaller rectangular prism

SA=2B+Ph

where

B is the area of the base

P is the perimeter of the base

h is the height of the prism

SA=2[(2.5)(1.5)]+2(2.5+1.5)(1.5)

SA=7.5+12=19.5\ m^2

step 2

Find the surface area of the cube

SA=6s^2

where

s is the length side of the cube

SA=6(2.5^2)=37.5\ m^2

step 3

Find the lateral area of the cylinder

LA=\pi DL

where

D is the diameter of the base

L is the length of cylinder

LA=(3.14)(0.5)(3.5)=5.495\ m^2

step 4

Find the area of the base of cylinder

A=\pi r^{2}

we have

r=0.5/2=0.25\ m ----> the radius is half the diameter

A=(3.14)(0.25)^{2}=0.19625\ m^2

step 5

Find the surface are of the composite figure

we know that

The surface area of the composite figure is equal to the surface area of the smaller rectangular prism plus the lateral area of the cylinder, plus the surface area of the cube, minus the area of the two bases of cylinder

so

SA=19.5+5.495+37.5-2(0.19625)=62.1\ m^2

5 0
3 years ago
I need help PLEASE!!!
Hitman42 [59]

Answer:

1. ∠A and ∠B are right angles.                      Given

2. m∠A = m∠ B                                               All right angles are congruent.

3. ∠BEC≅ ∠AED                                              Vertical angles are congruent

4. ΔCBE ~ ΔDAE                                               AA

Step-by-step explanation:

A proof always begins with the givens.

1. ∠A and ∠B are right angles. -------------->Given

2. m∠A = m∠ B  are equal since----------->  All right angles are congruent.

3. ∠BEC≅ ∠AED  are also equal since---->Vertical angles are congruent

4. ΔCBE ~ ΔDAE since two angles are equal----------> AA

7 0
3 years ago
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