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ziro4ka [17]
3 years ago
14

Numerical quantity for two dozen eggs

Mathematics
2 answers:
makvit [3.9K]3 years ago
5 0
Two dozen eggs are 24 eggs in numeral quantity
pentagon [3]3 years ago
4 0
One Dozen = 12
Two Dozen = 24
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PLEASE HELP ASAP 25 PTS + BRAINLIEST TO RIGHT/BEST ANSWER
MAVERICK [17]

Answer:

(-2,-5)

Step-by-step explanation:

y = 2x-1

3x-y = -1

We can use substitution

3x - (2x-1) = -1

Distribute the minus sign

3x -2x+1 = -1

x +1 =-1

Subtract 1 from each side

x+1-1 = -1-1

x=-2

Now we need to find y

y = 2x-1

y = 2(-2) -1

y = -4-1

y = -5

(-2,-5)

3 0
3 years ago
A student believes she can earn between $5200 and $6250 from her summer job. She knows that she will have to buy four new tires
ohaa [14]

Answer:

4180 to 5230$

Step-by-step explanation:

1. Lets find the expenses she has to make -> 4*90+660=1020$(4 new tires + other expenses)

2. As her income is not set strictly we can guess the range she can save(More accurate calculations depends on her strict incomes). We have 5200-1020=4180$(if she gets paid 5200$) and 6250-1020= 5230$(if she gets paid 6250$).

3. So the range of savings vary from 4180 to 5230$$.

4 0
3 years ago
Need help ASAP!!! Tysm if you answer!! IF YOU ANSWER RIGHT AND GET BRAINLIEST!!
Delvig [45]

Answer:

a_n = 3 ({5})^{n - 1}

Step-by-step explanation:

The given sequence is 3,15,75,375,...

The first term of this geometric sequence is

a_1=3

The common ratio is

r =  \frac{15}{3}  = 5

The explicit formula is given by:

a_n = a_1 {r}^{n - 1}

We plug the first term and common ratio into the formula to get:

a_n = 3 ({5})^{n - 1}

8 0
3 years ago
Suppose that X has a Poisson distribution with a mean of 64. Approximate the following probabilities. Round the answers to 4 dec
o-na [289]

Answer:

(a) The probability of the event (<em>X</em> > 84) is 0.007.

(b) The probability of the event (<em>X</em> < 64) is 0.483.

Step-by-step explanation:

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 64.

The probability mass function of a Poisson distribution is:

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

(a)

Compute the probability of the event (<em>X</em> > 84) as follows:

P (X > 84) = 1 - P (X ≤ 84)

                =1-\sum _{x=0}^{x=84}\frac{e^{-64}(64)^{x}}{x!}\\=1-[e^{-64}\sum _{x=0}^{x=84}\frac{(64)^{x}}{x!}]\\=1-[e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{84}}{84!}]]\\=1-0.99308\\=0.00692\\\approx0.007

Thus, the probability of the event (<em>X</em> > 84) is 0.007.

(b)

Compute the probability of the event (<em>X</em> < 64) as follows:

P (X < 64) = P (X = 0) + P (X = 1) + P (X = 2) + ... + P (X = 63)

                =\sum _{x=0}^{x=63}\frac{e^{-64}(64)^{x}}{x!}\\=e^{-64}\sum _{x=0}^{x=63}\frac{(64)^{x}}{x!}\\=e^{-64}[\frac{(64)^{0}}{0!}+\frac{(64)^{1}}{1!}+\frac{(64)^{2}}{2!}+...+\frac{(64)^{63}}{63!}]\\=0.48338\\\approx0.483

Thus, the probability of the event (<em>X</em> < 64) is 0.483.

5 0
3 years ago
A veterinary technician recorded the weights of 12 puppies. Of these 12 puppies, 3 weighed over 6 pounds. What percentage of the
OLga [1]

The percentage value of puppies weighing over 6 pounds is 25%

What is Percentage?

Percentage is a relative number that represents the hundredth part of any quantity. One percent (symbolised 1%) is one hundredth part; consequently, 100 percent denotes the complete amount and 200 percent defines twice the supplied amount.

Solution:

Given that there are 12 puppies in total and 3 of them weighted above 6 pounds

To calculate the percentage we need to divide 3 by 12 and then multiply it by 100

= (3/12) * 100

= 25%

To learn more about <u>Percentage</u> from the given link

brainly.com/question/843074

#SPJ1

3 0
1 year ago
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