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Alekssandra [29.7K]
2 years ago
6

Raymond bought 3/4 of a dozen rolls. How many rolls did he buy? Also, (another question) Jonathan rode 1.05 miles on Friday, 1.5

miles on Saturday, 1.25 mile son Monday, and 1.1 miles on Tuesday. On which day did he ride the shortest distance? 
Mathematics
2 answers:
alexgriva [62]2 years ago
7 0
Raymond bought 9 rolls and Jonathan rode Friday the least.
Gnom [1K]2 years ago
5 0
Answer to question one:
9 rolls
Answer to question two:
2.8
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Blue's Berry Farm charges Percy a total of
kaheart [24]

Answer:

The price for each kilogram of strawberries is $7.50

Step-by-step explanation:

<u><em>The question is</em></u>

Blues berry farm charges Percy a total of $24.75 for entrance and 2.5 kilograms of strawberries. The entrance fee is $6 and the price for each kilogram of strawberries is constant

Determine the price for each kilogram of strawberries

Let

x ----> the price for each kilogram of strawberries

we know that

The entrance fee plus the price of each kilogram of strawberries multiplied by the number of kilogram of strawberries must be equal to $24.75

so

The linear equation that represent this situation is

2.5x+6=24.75

solve for x

subtract 6 both sides

2.5x=24.75-6\\2.5x=18.75

divide by 2.5 both sides

x=\$7.50

therefore

The price for each kilogram of strawberries is $7.50

8 0
2 years ago
In ΔABC, AD and BE are the angle bisectors of ∠A and ∠B and DE ║ AB . If m∠ADE is with 34° smaller than m∠CAB, find the measures
Charra [1.4K]

Answer:

34°

Step-by-step explanation:

If m∠ADE is with 34° smaller than m∠CAB, then denote

m∠ADE=x°,

m∠CAB=(x+34)°.

Since  DE ║ AB, then

m∠ADE=m∠DAB=x°.

AD is angle A bisector, then

m∠EAD=m∠DAB=x°.

Thus,

m∠CAB=m∠CAD+m∠DAB=(x+x)°=2x°.

On the other hand,

m∠CAB=(x+34)°,

then

2x°=(x+34)°,

m∠ADE=x°=34°.

7 0
2 years ago
At a family reunion, each of Sana aunts and uncles is getting photographed once, The aunts are taking pictures in groups of5 and
yaroslaw [1]

This question is based on least common multiple method.

As given in the question,

Aunts are taking pictures in group of = 5

Uncles are taking pictures in group of = 10

So we have been asked what is the minimum number of aunts Sana have.

As we have been given that the number of aunts and uncles is equal so to find the minimum number of aunts we will apply the least common multiple method.

So we get LCM of 5,10 as 10

Hence there are minimum 10 aunts.


3 0
2 years ago
Read 2 more answers
Shayna enlarged a square photo by adding 10 inches to each side so it could be seen on a large poster. The area of the enlarged
Over [174]
The area of a square is expressed as the length of the side to the power of two, A = s^2. We were given the area of the enlarged photo which is 256 in^2. Also, it was stated that the length of the enlarged photo is the length of the original photo plus ten inches. So, from these statements we can make an equation to solve for x which represents the length of the original photo.

A = s^2
where s = (x+10)
A = (x+10)^2 = 256
Solving for x,
x= 6 in.

The dimensions of the original photo is 6 x 6.
8 0
2 years ago
Read 2 more answers
In 1898 L. J. Bortkiewicz published a book entitled The Law of Small Numbers. He used data collected over 20 years to show that
attashe74 [19]

Answer:

(a) The probability of more than one death in a corps in a year is 0.1252.

(b) The probability of no deaths in a corps over 7 years is 0.0130.

Step-by-step explanation:

Let <em>X</em> = number of soldiers killed by horse kicks in 1 year.

The random variable X\sim Poisson(\lambda = 0.62).

The probability function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,...

(a)

Compute the probability of more than one death in a corps in a year as follows:

P (X > 1) = 1 - P (X ≤ 1)

             = 1 - P (X = 0) - P (X = 1)

             =1-\frac{e^{-0.62}(0.62)^{0}}{0!}-\frac{e^{-0.62}(0.62)^{1}}{1!}\\=1-0.54335-0.33144\\=0.12521\\\approx0.1252

Thus, the probability of more than one death in a corps in a year is 0.1252.

(b)

The average deaths over 7 year period is: \lambda=7\times0.62=4.34

Compute the probability of no deaths in a corps over 7 years as follows:

P(X=0)=\frac{e^{-4.34}(4.34)^{0}}{0!}=0.01304\approx0.0130

Thus, the probability of no deaths in a corps over 7 years is 0.0130.

6 0
1 year ago
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