All categories

2 years ago

A flowerbed is 40 feet long it has a perimeter of 130 ft select true or false.The width of the flowerbed is90 ft

1 answer:

3 0

THIS IS FALSE!!
because if they give you that the perimeter is 130 and the flower bed is 40 feet long we would subtract 40 from 130 twice. lets do the math:)
130 - 40 - 40 = 50
and in order to find our width we divide 50 by 2. lets do the math:)
50 / 2 = 25!!
so our width here is 25 not 90
have a good one:)

You might be interested in

a box holds nine smaller boxes. each of the smaller boxes holds nine plastic containers. each plastic contai.er holds nine bags.

svp [43]

So, basically ..you just multiply
9*9*9*9
6561

4 0

2 years ago

Food allergies affect an estimated 7% of children under age 5 in the US. What is the probability that in a kindergarden class of

bezimeni [28]

Answer:

Probability of atleast one of 12 student has food allergies ≈ 0.58 ( approx)

Step-by-step explanation:

Given: Probability of a children under age 5 has food allergies = 7%

= $\frac{7}{100}$

To find : Probability of atleast one of 12 student has food allergies

Probability of a chindren under age 5 does not have food allergies = $1-\frac{7}{100}$

⇒ prob = $\frac{93}{100}$

now we find Probability of atleast one of 12 student has food allergies this means we have to find prob of 1 student, 2 student, 3 student, till 12 student have allergy out of 12 student of class then add all prob.

But instead of finding all these probability we find probability of student having no allergy.i.e., 0 student then subtract it from 1(total probability)

Probability of 0 student having allergy out of 12 student = $(\frac{93}{100})^{12}$

Therefore, Probability of atleast one of 12 student has food allergies

= $1-(\frac{93}{100})^{12}$

= $0.581403702521$

≈ 0.58 ( approx)

8 0

2 years ago

Read 2 more answers

Two production lines are used to pack sugar into 5 kg bags. Line 1 produces twice as many bags as does line 2. One percent of ba

enyata [817]

Answer:

P(Bag is Defective) = 0.0167

Step-by-step explanation:

Line 1 produces twice as many bags as line 2. Let x be the number of bags produced by line 2.

No. of bags produced by line 2 = x

No. of bags produced by line 1 = 2x

Probability that the bag has been produced by line 1 can be written as:

P(Line 1) = No. of bags produced by line 1/Total no. of bags

= 2x/(x+2x)

= 2x/3x

P(Line 1) = 2/3. Similarly,

P(Line 2) = x/3x

P(Line 2) = 1/3

1% bags produced by line 1 are defective so the probability of line 1 producing a defective bag is:

P(Defective|Line 1) = 0.01

3% of bags from line 2 are defective, so:

P(Defective|Line 2) = 0.03

b. The probability that the chosen bag is defective can be calculated through the conditional probability formula:

P(A|B) = P(A∩B)/P(B)

P(A∩B) = P(A|B)*P(B)

The chosen defective bag can be either from line 1 or from line 2. So, the probability that the chosen bag is defective is:

P(Bag is Defective) = P(Defective and from Line 1) + P(Defective and from Line 2)

= P(D∩Line 1) + P(D∩Line 2)

= P(Defective|Line 1)*P(Line 1) + P(Defective|Line 2)*P(Line 2)

= (0.01)*(2/3) + (0.03)(1/3)

P(Bag is Defective) = 0.0167

7 0

2 years ago

Read 2 more answers

Markus scored 85, 92, 82, and 94 on the first four tests of the semester. His teacher has not yet told him his score on his fift

algol13

It’s x=83 I believe

5 0

2 years ago

Read 2 more answers

The weight of corn chips dispensed into a 14-ounce bag by the dispensing machine has been identified as possessing a normal dist

mamaluj [8]

The probability that a normally distributed dataset with a mean, μ, and statndard deviation, σ, exceeds a value x, is given by

$P(X\ \textgreater \ x)=1-P(X\ \textless \ x)=1-P\left(z\ \textless \ \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \right)$

Given that the weight of corn chips dispensed into a 14-ounce bag by the dispensing machine is a normal distribution with a mean of 14.5 ounces and a standard deviation of 0.2 ounce.

If 100 bags of chips are randomly selected the probability that the mean weight of these 100 bags exceeds 14.6 ounces is given by

$P(X\ \textgreater \ 14.6)=1-P\left(z\ \textless \ \frac{14.6-14.5}{\frac{0.2}{\sqrt{100}}} \right) \\ \\ =1-P\left(z\ \textless \ \frac{0.1}{\frac{0.2}{10}} \right)=1-P\left(z\ \textless \ \frac{0.1}{0.02} \right) \\ \\ =1-P(z\ \textless \ 5)=1-1=0$

Therefore, the probability that the mean weight of these 100 bags exceeds 14.6 ounces is 0.

7 0

2 years ago