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2 years ago

A cake for 6 people requires 4 eggs.How many eggs are needed to make a cake big enough for 9 people

Mathematics

1 answer:

Sphinxa [80]2 years ago

7 0

Answer:

Step-by-step explanation:

Divide the amount of eggs by the amount of people this gives you the amount of eggs needed for one person then times by 9

(4÷6) x9

You might be interested in

The probability that Scott will win his next tennis match is 3/5

goblinko [34]

Answer:

the probability that he will not win is 2/5 cause it's just the chance that he doesn't win

6 0

2 years ago

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and tes

DerKrebs [107]

Answer:

a) The probability that this whole shipment will be accepted is 30%.

b) Many of the shipments with this rate of defective aspirin tablets will be rejected.

Step-by-step explanation:

We have a shipment of 3000 aspirin tablets, with a 5% rate of defects.

We select a sample of size 48 and test for defectives.

If more than one aspirin is defective, the batch is rejected.

The amount of defective aspirin tablets X can be modeled as a binomial distribution random variable, with p=0.55 and n=48

We have to calculate the probabilities that X is equal or less than 1: P(X≤1).

$P(X\leq1)=P(X=0)+P(X=1)\\\\\\P(0)=\binom{48}{0}(0.05)^0(0.95)^{48}=1*1*0.0853=0.0853\\\\\\P(1)=\binom{48}{1}(0.05)^1(0.95)^{47}=48*0.05*0.0897=0.2154\\\\\\P(X\eq1)=0.0853+0.2154=0.3007$

8 0

2 years ago

QUESTION 7 Solve for x. (3x + 32) (5x-8) O a. 3 b. 12 O c. 20 O d. 19.5

Alchen [17]

Answer:

The answer is x = 20

Step-by-step explanation:

If you set both equations equal to each other and solve for x, you will get 20.

3x+32 = 5x-8 add 8 to both sides

3x+40 = 5x subtract 3x from both sides

40 = 2x divide by 2 on both sides

20 = x

7 0

2 years ago

i) Baraka bought ksh.20,000 worth of shares of a company each month. During the first three months, he bought the shares at a pr

Sladkaya [172]

The average price paid by him for the shares after 3 months is ksh. 163.33

<h3>Average</h3>

Total value of shares bought = ksh.20,000

Amount of shares bought in the first three months = ksh.120, ksh.160 and ksh.210

Average price paid for the shares after 3 months

= (120 + 160 + 210) / 3

= 490 / 3

= 163.333333333333

Approximately,

ksh. 163.33

Learn more about average:

brainly.com/question/20118982

#SPJ1

4 0

1 year ago

What is Limit of StartFraction StartRoot x + 1 EndRoot minus 2 Over x minus 3 EndFraction as x approaches 3?

scoray [572]

Answer:

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \boxed{ \frac{1}{4} }$

General Formulas and Concepts:

Calculus

Limits

Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to c} x = c$

Special Limit Rule [L’Hopital’s Rule]:
$\displaystyle \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$

Differentiation

Derivatives
Derivative Notation

Derivative Property [Addition/Subtraction]:
$\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$
Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:
$\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify given limit.

$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3}$

Step 2: Find Limit

Let's start out by directly evaluating the limit:

[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} = \frac{\sqrt{3 + 1} - 2}{3 - 3}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \frac{\sqrt{3 + 1} - 2}{3 - 3} \\& = \frac{0}{0} \leftarrow \\\end{aligned}$

When we do evaluate the limit directly, we end up with an indeterminant form. We can now use L' Hopital's Rule to simply the limit:

[Limit] Apply Limit Rule [L' Hopital's Rule]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\\end{aligned}$
[Limit] Differentiate [Derivative Rules and Properties]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \leftarrow \\\end{aligned}$
[Limit] Apply Limit Rule [Variable Direct Substitution]:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \leftarrow \\\end{aligned}$
Evaluate:
$\displaystyle \begin{aligned}\lim_{x \to 3} \frac{\sqrt{x + 1} - 2}{x - 3} & = \lim_{x \to 3} \frac{(\sqrt{x + 1} - 2)'}{(x - 3)'} \\& = \lim_{x \to 3} \frac{1}{2\sqrt{x + 1}} \\& = \frac{1}{2\sqrt{3 + 1}} \\& = \boxed{ \frac{1}{4} } \\\end{aligned}$