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2 years ago

Debbie used 40% of her flour for cookies she started with 24 cups how many cups did debbie use?

Mathematics

2 answers:

Lorico [155]2 years ago

7 0

the answer is Lil Debbie™ 9.6 cup used.

shtirl [24]2 years ago

6 0

Answer:

24 x 40% = 9.6

Step-by-step explanation:

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Denise wants to cover a 21.75 square foot wall with weathered wood. If each box of weathered wood contains 1.5 square feet of wo

natka813 [3]

Answer: 15 boxes

Step-by-step explanation:

From the question, we are informed that Denise wants to cover a 21.75 square foot wall with weathered wood and that each box of weathered wood contains 1.5 square feet of wood.

The number of boxes of weathered wood that she need will be calculated by dividing 21.75 square feet by 1.5 square feet. This will be:

= 21.75/1.5

= 14.5

= 15 boxes approximately

6 0

3 years ago

Let X be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodo

Shkiper50 [21]

Answer:

a) $P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.6288$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.1954=0.4335$

b) $P(4\leq X\leq 8)=0.1954+0.1563+0.1042+0.0595+0.0298=0.5452$

c) $P(X \geq 8) = 1-P(X$

d) $P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

Step-by-step explanation:

Let X the random variable that represent the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. We know that $X \sim Poisson(\lambda=4)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda=4$ , $Var(X)=\lambda=2$ , $Sd(X)=2$

a. Compute both P(X≤4) and P(X<4).

$P(X\leq 4)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)+P(X=4)$

Using the pmf we can find the individual probabilities like this:

$P(X=0)=\frac{e^{-4} 4^0}{0!}=e^{-4}=0.0183$

$P(X=1)=\frac{e^{-4} 4^1}{1!}=0.0733$

$P(X=2)=\frac{e^{-4} 4^2}{2!}=0.1465$

$P(X=3)=\frac{e^{-4} 4^3}{3!}=0.1954$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X\leq 4)=0.0183+0.0733+ 0.1465+0.1954+0.1954=0.9646$

$P(X< 4)=P(X\leq 3)=P(X=0)+P(X=1)+ P(X=2)+P(X=3)$

$P(X< 4)=P(X\leq 3)=0.0183+0.0733+ 0.1465+0.5311=0.7692$

b. Compute P(4≤X≤ 8).

$P(4\leq X\leq 8)=P(X=4)+P(X=5)+ P(X=6)+P(X=7)+P(X=8)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(X=7)=\frac{e^{-4} 4^7}{7!}=0.0595$

$P(X=8)=\frac{e^{-4} 4^8}{8!}=0.0298$

$P(4\leq X\leq 8)=0.1954+0.1563+ 0.1042+0.0595+0.0298=0.5452$

c. Compute P(8≤ X).

$P(X \geq 8) = 1-P(X$

d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

The mean is 4 and the deviation is 2, so we want this probability

$P(4\leq X \leq 6)=P(X=4)+P(X=5)+P(X=6)$

$P(X=4)=\frac{e^{-4} 4^4}{4!}=0.1954$

$P(X=5)=\frac{e^{-4} 4^5}{5!}=0.1563$

$P(X=6)=\frac{e^{-4} 4^6}{6!}=0.1042$

$P(4\leq X \leq 6)=0.1954+0.1563+0.1042=0.4559$

4 0

3 years ago

Janice is making a soup recipe that normally yields 8 servings. The recipe calls for 3 pounds of chicken and a certain number of

kondaur [170]

Answer:Janice needs 5 pounds of vegetables.

Step-by-step explanation:

5 0

3 years ago

Let p(x)=3x-7. If p(x)=0,find x. A. -7

Misha Larkins [42]

Answer:

Step-by-step explanation:

0=3x-7

0=3x+7

7=3x/3

7/3=x

4 0

3 years ago

The rectangle below has an area of 6n^4+20n^3+14n^26n 4 +20n 3 +14n 2 6, n, start superscript, 4, end superscript, plus, 20, n,

Sindrei [870]

Given:

Area of rectangle = $6n^4+20n^3+14n^2$

Width of the rectangle is equal to the greatest common monomial factor of $6n^4, 20n^3,14n^2$ .

To find:

Length and width of the rectangle.

Solution:

Width of the rectangle is equal to the greatest common monomial factor of $6n^4, 20n^3,14n^2$ is

$6n^4=2\times 3\times n\times n\times n\times n$

$20n^3=2\times 2\times 5\times n\times n\times n$

$14n^2=2\times 7\times n\times n$

Now,

$GCF(6n^4, 20n^3,14n^2)=2\times n\times n=2n^2$

So, width of the rectangle is $2n^2$ .

Area of rectangle is

$Area=6n^4+20n^3+14n^2$

Taking out GCF, we get

$Area=2n^2(3n^2+10n+7)$

We know that, area of a rectangle is the product of its length and width.

Since, width of the rectangle is $2n^2$ , therefore length of the rectangle is $(3n^2+10n+7)$ .

5 0

3 years ago