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2 years ago

Ciaran scores 76% work out 76% of 75 marks Help pls

1 answer:

3 0

To get 76% of 75 you turn the fraction to a decimal, then multiply it to the number. 76% as a decimal us 0.76 and 0.76 * 75 = 57

So, Ciaran got 57 out of 75 questions right.

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What is the value of x in the diagram?

DerKrebs [107]

The answer is 27 (hope this helps you)

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2 years ago

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What is the area of the following trapezoid?<br> 13.5 yd 2<br> 3.38 yd 2<br> 6.75 yd 2<br> 8.25 yd 2

Neporo4naja [7]

Answer:

$area = \frac{a + b}{2} h \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{2 + 7}{2} \times 1.5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6.75 {yd}^{2}$

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2 years ago

Part A: Xavier needs to collect at least 120 cans for a food drive to earn community service credit. He has already collected 64

Sergeu [11.5K]

Answer:

Step-by-step explanation:

He needs to collect at least 120 so if c is what he must still collect

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1 year ago

Suppose that there are five young women and six young men on an island. Each woman is willing to marry some of the men on the is

pochemuha

Answer:

Based on the current information, the last man remaining unmarried will be a choice between two

Jason marries Anna

Matt marries Elizabeth

Kevin marries Barbara

Larry marries Diane

Carol can marry either Nick or Oscar

Step-by-step explanation:

The basic method to sort which woman will be marrying which man is simple.

We have to extract the choice by comparing choices of each woman side by side and then selecting the least common one and hence sorting one by one.

First, let's name all the women and men

1. Anna

2. Barbara

3. Carol

4. Diane

5. Elizabeth

We will refer to them with the initial letter of their name . A for Anna, B for Barbara, C for Carol, D for Diane, E for Elizabeth.

The men :

1. Jason

2. Larry

3. Matt

4. Kevin

5. Oscar

6. Nick

And they will be referred as J for Jason, L for Larry, M for Matt, K for Kevin, O for Oscar and N for Nick.

Options for A : J , L ,M

Options for B : K, L

Options for C : J, N , O

Options for D : J, L , N , O

Options for E : J, M

As we can see , no one else wants Kevin other than Barbara so :

Kevin marries Barbara

Moving onwards:

Elizabeth can marry either Jason or Matt,

Meanwhile Anna can marry either Jason , Larry or Matt

Hence,

Matt marries Eilzabeth

as Anna has two more choices still which are Jason and Larry.

Further, seeing the options for Carol and Diane

Carol can marry either Jason, Nick or Oscar while Diane can marry Larry as well as Jason , Nick and Oscar

We can select Jason as the match for Anna as Larry is also a choice for Diane.

Hence, Jason marries Anna

Now we are left with Larry , Nick and Oscar.

Since Carol can only decide from Nick and Oscar but Diane can decide from Larry, Nick and Oscar. Larry is the ideal option for Diane

so, Larry marries Diane

Now in the end, Carol can either choose between Nick or Oscar.

If she chooses Nick to marry, Oscar is left single

If she chooses Oscar to marry, Nick is left single.

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2 years ago

Simplify what is written here

Inessa05 [86]

When multiplying terms that have the same base (in this case, $x$ ) but different exponents, you can add their exponents to make them one term.

In this case, we can add the exponents of $x$ :

$x \cdot x^5 \cdot x^5$

$x^{(1 + 5 + 5)}$

$\boxed{x^{11}}$

A simplified version is $x^{11}$ .

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2 years ago

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