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3 years ago

The Gcf of 6,14,18,and the Lcd of 5,9,13

Mathematics

1 answer:

mrs_skeptik [129]3 years ago

6 0

Gcf of 6,14,18: 2
And you mean lcm?

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800 g pineapple

neonofarm [45]

Answer:

30 people

Step-by-step explanation:

The question is as following:

Here are the ingredients for pineapple sherbet for 6 people

800g pineapple 4 egg whites 1/2 lemon 100g caster sugar

Dan makes pineapple sherbet he uses 2 1/2 lemons

work out how many people he makes pineapple sherbet for?

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

1/2 lemon needed for pineapple sherbet for 6 people

2 1/2 lemons needed for pineapple sherbet for x people

1/2 : 6 = 2 1/2 : x

∴ x = ( 6 * 2 1/2 ) ÷ (1/2)

x = 6 * 5/2 * 2 = 30 people

So, he make the pineapple sherbet for 30 people

5 0

2 years ago

15/4 in to equal mixed number

Temka [501]

9514 1404 393

Answer:

3 3/4

Step-by-step explanation:

$\dfrac{15}{4}=\dfrac{12+3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\\\\=3+\dfrac{3}{4}=\boxed{3\dfrac{3}{4}}$

The integer part of the mixed number is the quotient when the numerator is divided by the denominator. The remainder from that division is the numerator of the fraction.

15/4 = 15 ÷ 4 = 3 r 3 = 3 3/4

5 0

2 years ago

Suppose that an airline uses a seat width of 16.5 in. Assume men have hip breadths that are normally distributed with a mean of

Alexxx [7]

Answer:

a) 0.018

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 14.4 in

Standard Deviation, σ = 1 in

We are given that the distribution of breadths is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(breadth will be greater than 16.5 in)

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5 - 14.4}{1}) = P(z > 2.1)$

$= 1 - P(z \leq 2.1)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 0.982 = 0.018 = 1.8\%$

0.018 is the probability that if an individual man is randomly selected, his hip breadth will be greater than 16.5 in.

b) P( with 123 randomly selected men, these men have a mean hip breadth greater than 16.5 in)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

P(x > 16.5)

$P( x > 16.5) = P( z > \displaystyle\frac{16.5-14.4}{\frac{1}{\sqrt{123}}}) = P(z > 23.29)$

$= 1 - P(z \leq 23.29)$

Calculation the value from standard normal z table, we have,

$P(x > 16.5) = 1 - 1 = 0$

There is 0 probability that 123 randomly selected men have a mean hip breadth greater than 16.5 in

4 0

3 years ago

Meg found another treasure between -17 feet and -2 1/2 feet. Where was that treasure?

Inessa [10]

The answer is -19 1/2 feet

3 0

3 years ago

A car dealership pays a wholesome price of $12,000 to purchase a vehicle. The car dealership wants to make a 32% profit. by how

katen-ka-za [31]

Multiply $12,000 with .32. Then add the answer to $12,000

6 0

3 years ago