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3 years ago

Three eggs are required to make two dozen muffins how many eggs are needed to make 12 dozen muffins

Mathematics

1 answer:

nordsb [41]3 years ago

3 0

1 and a half !!!!!!!!!!!!!!!

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Which expression represents the area of this figure?

dalvyx [7]

Answer:Where is the figure?

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3 years ago

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HELPP ASAP!!step by step anwserssw

Natasha_Volkova [10]

X - The total cost
3 persons
Below $120 p/p
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120 * 3 = 360
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3 years ago

NEED HELP WORTH 100 POINTS SHOW YOUR WORK PLEASE

tia_tia [17]

Answer:

Q1:

$-5\sqrt{27c^2}$

$-5\sqrt{27}\sqrt{c^2}$

$-5\times\sqrt{9} \sqrt{3} \sqrt{c^2}$

$-5\times3\sqrt{3}\sqrt{c^2}$

$-5\times3\sqrt{3}c$

$-15\sqrt{3}c$

Q2:

$5\sqrt{72}-3\sqrt{32}$

$5\sqrt{36} \sqrt{2} -3\sqrt{16} \sqrt{2}$

$5\times6\sqrt{2}-3\times4\sqrt{2}$

$30\sqrt{2}-12\sqrt{2}$

$18\sqrt{2}$

Q3:

$\sqrt{108yz}+3\sqrt{98yz}+2\sqrt{75yz}$

$\sqrt{36} \sqrt{3} \sqrt{yz} +3\sqrt{49} \sqrt{2} \sqrt{yz} +2\sqrt{25} \sqrt{3} \sqrt{yz}$

$6\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}+10\sqrt{3}\sqrt{yz}$

$16\sqrt{3}\sqrt{yz}+21\sqrt{2}\sqrt{yz}$

Q4:

$\sqrt{15n^2}\sqrt{10n^3}$

$\sqrt{15}n\sqrt{10n^3}$

$\sqrt{15}n\sqrt{10}\sqrt{n^2}\sqrt{n}$

$\sqrt{15}\sqrt{10}n^2\sqrt{n}$

$\sqrt{5}\sqrt{3}\sqrt{5}\sqrt{2}n^2\sqrt{n}$

$5\sqrt{3}\sqrt{2}n^2\sqrt{n}$

$5\sqrt{6}n^2\sqrt{n}$

Q5:

$\frac{\sqrt{3x^2y^3}}{4\sqrt{5xy^3}}$

$\frac{\sqrt{3}xy\sqrt{y}}{4\sqrt{5}y\sqrt{x}\sqrt{y}}$

Cancel off $\sqrt{x}$ :

$\frac{\sqrt{3}y\sqrt{x}\sqrt{y}}{4\sqrt{5}y\sqrt{y}}$

Cancel off $y:$

$\frac{\sqrt{3}\sqrt{x}\sqrt{y}}{4\sqrt{5}\sqrt{y}}$

Cancel off $\sqrt{y} :$

$\frac{\sqrt{3}\sqrt{x}}{4\sqrt{5}}$

Multiply $\sqrt{5}$ on the numerator and denominator:

$\frac{\sqrt{3}\sqrt{x}\sqrt{5}}{4\sqrt{5}\sqrt{5}}$

$\frac{\sqrt{15}\sqrt{x}}{20}$

5 0

3 years ago

g 78% of all Millennials drink Starbucks coffee at least once a week. Suppose a random sample of 50 Millennials will be selected

Tatiana [17]

Answer:

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

We know that n = 50 and p =0.78.

We need to check the conditions in order to use the normal approximation.

$np=50*0.78=39 \geq 10$

$n(1-p)=50*(1-0.78)=11 \geq 10$

Since both conditions are satisfied we can use the normal approximation and the distribution for the proportion is given by:

$p \sim N (p, \sqrt{\frac{p(1-p)}{n}})$

With the following parameters:

$\mu_ p = 0.78$

$\sigma_p = \sqrt{\frac{0.78*(1-0.78)}{50}}= 0.0586$

6 0

3 years ago

HELP!!!!! U GOTTA DO THIS PLS

salantis [7]

Answer:

60 almonds?

Step-by-step explanation:

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3 years ago

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