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3 years ago

At a restaurant the ratio of kids meals sold to adult meals sold was 7:3. If

Mathematics

1 answer:

Bezzdna [24]3 years ago

4 0

Answer:

Step-by-step explanation:

because 35 divided by 5 is 7 and 15 divided by 15 is 3 making 7:3

hope it helps

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Is 6+2 +4x equivalent to 6- ( 2-4x) . if not , why not ?

Xelga [282]

Well lets solve for both.
6 + 2 + 4x
Combine like terms.
8 + 4x
Now solve the other one.
6 - (2 - 4x)
Distribute.
6 - 2 + 4x
Combine like terms.
4 + 4x.
These equations are not equivalent because instead of adding two, like in the first one, the second problem is subtracting. So you get 8 + 4x and 4 + 4x. I hope this helps love! :)

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3 years ago

What is the lowest common multiple of 120 160 180

Anna11 [10]

20 it the least common multiple of the numbers given

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3 years ago

Jason wrote 1 1/3 as a sum of 3 fractions. None of the fractions had a denominator of 3. What fractions might Jason have used?

ziro4ka [17]

11/6 , 1/6 and 10/6 are the answer

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3 years ago

create a divison problem with a 2-digit divisor using more than 1 set of compatible numbers, observe what happen when you estima

liubo4ka [24]

Here Is A Dividend That Might Help Is 12÷6=2

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3 years ago

Suppose that 15% of the fields in a given agricultural area are infested with the sweet potato whitefly. One hundred fields in t

hram777 [196]

Answer:

a. $\mu=15$

b. $\mu=7.8586 \ and \ \mu=22.1414$

c. Choice A- Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

Step-by-step explanation:

a.Binomial distribution is defined by the expression

$P(X=k)=C_k^n.p^k.(1-p)^{n-k}$

Let n be the number of trials, $n=100$

and p be the probability of success, $p=15\%$

The mean of a binomial distribution is the probability x sample size.

$\mu=np=100\times0.15=15$

b.Limits within which p is approximately 95%

sd of a binomial distribution is given as: $\sigma=\sqrt npq\\q=1-p$

Therefore, $\sigma=\sqrt(100\times0.015\times0.85)=3.5707$

Use the empirical rule to find the limits. From the rule, approximately 95% of the observations are within to standard deviations from mean.

$sd_1=>\mu-2\sigma=15-2\times3.3507=7.8586\\sd_2=>\mu-2\sigma=15+2\times3.3507=22.1414$

Hence, approximately 95% of the observations are within 7.8586 and 22.1414 (areas of infestation).

c. $x=45$ is not within the limits in b above (7.8586,22.1414). X=45 appears to be a large area of infestation. A.Based on limits above, it is unlikely that we would see x = 45, so it might be possible that the trials are not independent.

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3 years ago