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uranmaximum [27]
4 years ago
7

Parker was picking shamroks he picked a total of 40 shamroks two faiths of the shamroks had four leaves the rest of the shamroks

had three leaves how many shamroks had three leaves
Mathematics
1 answer:
Gnesinka [82]4 years ago
3 0

Answer:

24 shamrocks had three leaves.

Step-by-step explanation:

We have been given that Parker was picking shamrocks he picked a total of 40 shamrocks two fifths of the shamrocks had four leaves. The rest of the shamrocks had three leaves.

First of all, we will find two fifths of 40 to find number of shamrocks with four leaves.

\text{Shamrocks with four leaves}=40\times \frac{2}{5}

\text{Shamrocks with four leaves}=8\times \frac{2}{1}

\text{Shamrocks with four leaves}=16

Now, we will subtract number of shamrocks with 4 leaves from total number of shamrocks to find number of shamrocks with 3 leaves as:

\text{Shamrocks with three leaves}=\text{Total shamrocks}-\text{Shamrocks with four leaves}

\text{Shamrocks with three leaves}=40-16

\text{Shamrocks with three leaves}=24

Therefore, 24 shamrocks had three leaves.

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Reed made a lasagna for dinner. That night, he ate1/4
Snezhnost [94]

Answer: 11/12

Step-by-step explanation:

First find the LCM of 4 and 3(12).  Then make the denominator of both fractions 12(3/12 and 8/12).  Then add the fractions to get that they ate 11/2 of the lasagna.

Hope it helps <3

6 0
4 years ago
There are 7 tigers in a castle,each tiger has 7 children,each child has 7 tiger friends,then how many tigers are there in the ca
Ksenya-84 [330]
There are 7 parent tigers
Parents = 7


7 tigers each have 7 children. That means there's total of 7 × 7 children.
Children = 7 × 7 = 49

Each child has 7 friends. That means there's total of 49 × 7 friends.
Friends = 49 × 7 = 343

Total = 7 + 49 +343 = 399 TIGERS
I'm not a tiger so I'm not counting myself.

3 0
4 years ago
A TEACHER IS CONSTRUCTING A MATHEMATICS TEST
Nuetrik [128]

The number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

<h3>How to determine the combination</h3>

Note that the formula for combination is given as;

Combination = \frac{n!}{r!(n-r)!}

From the information we have that;

There are 28 questions in the pool

The test should have a total of 10 questions;

6 difficult , 10 average and 12 easy questions

We are asked to determine the combination of;

3 difficult questions

4 average questions

3 easy questions

6C3 = \frac{6!}{3!(6-3)!}

6C3 = \frac{720}{36}

6C3 = 20

10C4 = \frac{10!}{4!(10-4)!}

10C4 = \frac{3628800}{17280}

10C4 = 210

12C4 = \frac{12!}{4!(12-4)!}

12C4 = 220

The number of ways of constructing the questions is

= 20 × 210 × 220

= 924, 000 ways

Thus, the number of ways of constructing questions from the pool of 28 questions if her test is to have 3 difficult, 4 average and 3 easy questions is  924, 000 ways

Learn more about combination here:

brainly.com/question/4658834

#SPJ1

7 0
2 years ago
I need help with this problem for my class! thanks
PSYCHO15rus [73]

Answer:

point on line= (0,8)

slope=4

7 0
3 years ago
Read 2 more answers
Help me find the volume of the composite figure
Butoxors [25]

Answer:

V = 97.911in^3

Step-by-step explanation:

Given

Shapes: Cube and Cone

Required

Determine the volume

First we calculate the volume of the cube

V_1 = l^3

Where

l = side length = 5.1

V_1 = 5.1^3

V_1 = 132.651

Next, calculate the volume of the cone using:

V_2 = \frac{\pi r^2h}{3}

Where

h = 5.1

r = \frac{1}{2} * Diameter

r = \frac{1}{2} * 5.1

r = 2.55

So, we have:

V_2 = \frac{\pi r^2h}{3}

V_2 = \frac{\pi * 2.55^2 * 5.1}{3}

V_2 = \frac{22 * 2.55^2 * 5.1}{7*3}

V_2 = \frac{729.5805}{21}

V_2 = 34.74

The volume of the figure is:

V=V_1 - V_2

V_1 = 132.651

V = 132.651 - 34.74

V = 97.911in^3

8 0
3 years ago
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