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3 years ago

Nancy needs 4/5 of a foot of ribbon to make a bow and she has 4 feet ribbon how many bows can she make

Mathematics

1 answer:

vlabodo [156]3 years ago

5 0

Answer:

I believe she can make one bow

Step-by-step explanation:

Iso if she has 5 feet of ribbon and only takes four feet of it then that leaves one foot left but if she has a perfect four foot then she can make a ribbon without extra

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Write the equation of the line.

Oksana_A [137]

Y = 1/3x - 2.
If you sub in 0, you’ll get (0, -2) which is the y intercept.

4 0

3 years ago

Read 2 more answers

Find the area of the composite figure.

Roman55 [17]

Answer:

The Area of the composite figure would be 76.26 in^2

Step-by-step explanation:

According to the Figure Given:

Total Horizontal Distance = 14 in

Length = 6 in

To Find :

The Area of the composite figure

Solution:

Firstly we need to find the area of Rectangular part.

So We know that,

$\boxed{ \rm \: Area \: of \: Rectangle = Length×Breadth}$

Here, Length is 6 in but the breadth is unknown.

To Find out the breadth, we’ll use this formula:

$\boxed{\rm \: Breadth = total \: distance - Radius}$

According to the Figure, we can see one side of a rectangle and radius of the circle are common, hence,

$\longrightarrow\rm \: Length \: of \: the \: circle = Radius$

Since Length = 6 in ;

$\longrightarrow \rm \: 6 \: in = radius$

Hence Radius is 6 in.

So Substitute the value of Total distance and Radius:

Total Horizontal Distance= 14
Radius = 6

$\longrightarrow\rm \: Breadth = 14-6$

$\longrightarrow\rm \: Breadth = 8 \: in$

Hence, the Breadth is 8 in.

Then, Substitute the values of Length and Breadth in the formula of Rectangle :

Length = 6
Breadth = 8

$\longrightarrow\rm \: Area \: of \: Rectangle = 6 \times 8$

$\longrightarrow \rm \: Area \: of \: Rectangle = 48 \: in {}^{2}$

Then, We need to find the area of Quarter circle :

We know that,

$\boxed{\rm Area_{(Quarter \; Circle) } = \cfrac{\pi{r} {}^{2} }{4}}$

Now Substitute their values:

r = radius = 6
π = 3.14

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 6 {}^{2} }{4}$

Solve it.

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times 36}{4}$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = \cfrac{3.14 \times \cancel{{36} } \: ^{9} }{ \cancel4}$

$\longrightarrow\rm Area_{(Quarter \; Circle)} =3.14 \times 9$

$\longrightarrow\rm Area_{(Quarter \; Circle) } = 28.26 \: {in}^{2}$

Now we can Find out the total Area of composite figure:

We know that,

$\boxed{ \rm \: Area_{(Composite Figure)} =Area_{(rectangle)}+ Area_{ (Quarter Circle)}}$

So Substitute their values:

$\rm Area_{(rectangle)}$ = 48
$\rm Area_{(Quarter Circle)}$ = 28.26

$\longrightarrow \rm \: Area_{(Composite Figure)} =48 + 28 .26$

Solve it.

$\longrightarrow \rm \: Area_{(Composite Figure)} =\boxed{\tt 76.26 \:\rm in {}^{2}}$

Hence, the area of the composite figure would be 76.26 in² or 76.26 sq. in.

$\rule{225pt}{2pt}$

I hope this helps!

3 0

2 years ago

gabriel put 6000 in a 2 year CD paying 4% interest, compounded monthly. After 2 years , he withdrew all his money. what was the

shutvik [7]

amount after 2 years = 6000(1 + (0.04/12))^24 = 6498.86

7 0

3 years ago

Read 2 more answers

I’m having a hard time understanding substitution. Please help!

Vikentia [17]

$b_1=\frac{2A}{h}-b_2$

Step-by-step explanation:

Given expression is;

$A=\frac{h(b_1+b_2)}{2}$

Solving for $b_1$

Multiplying both sides by 2

$2*A=\frac{h(b_1+b_2)}{2}*2\\2A=h(b_1+b_2)$

Dividing both sides by h

$\frac{2A}{h}=\frac{h(b_1+b_2)}{h}\\\frac{2A}{h}=b_1+b_2$

Subtracting $b_2$ from both sides

$\frac{2A}{h}-b_2=b_1+b_2-b_2\\\frac{2A}{h}-b_2=b_1\\b_1=\frac{2A}{h}-b_2$

$b_1=\frac{2A}{h}-b_2$

Keywords: division, subtraction

Learn more about subtraction at:

#LearnwithBrainly

3 0

3 years ago

A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from thi

SVETLANKA909090 [29]

Complete question:

Consider a class of 20 students consisting of 5 sophomores, 8 juniors, and 7 seniors. A math instructor assigns a group project in each of the scenarios below, the instructor selects 5 consecutive students from this class, keeping track (in order) of the level of the student that he gets.

1a. How many possible outcomes are in the sample space S? (An outcome is a 5- tuple of students.)

Let A2 denote the event that exactly 2 of the selected students are sophomore, A5 denote the event that exactly 5 of the selected students are the juniors, and A4 denote the event that exactly 4 of the selected students are seniors.

1b. Find the probabilities of each of these three events. A2, A5, A4

1c. Do the events A2, A5, A4 constitute a partition of the sample space?

Answer:

Given:

n = 20

a) The possible outcome in the sample space,S, =

ⁿCₓ = ²⁰C₅

$= \frac{20!}{(20-5)! 5!)}$