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Mrrafil [7]
4 years ago
14

How do I solve 13.65=h+4.88 it’s stupid I know but I need help!!

Mathematics
1 answer:
murzikaleks [220]4 years ago
4 0

Answer:

8.77

Step-by-step explanation:

13.65 - 4.88 = 8.77

8.77 = h

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12x + 45 < 117

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Each of forty-one pet owners was asked, "Does your pet eat more dry food or wet food?" Here are the results. Among the cat owner
Amiraneli [1.4K]

Answer: No, pet does not eat more dry food than wet food.

Step-by-step explanation:

Since we have given that

In case of cat owner,

Number who chose Dry food = 6

Number who chose Wet food = 8

In case of dog owners,

Number who chose Dry food = 14

Number who wet food = 13

So,Total number who chose Dry food = 6+14=20

Total number who chose Wet food = 8+13=21

So, the answer for this question ""Does your pet eat more dry food or wet food?" is No.

Hence, No, pet does not eat more dry food than wet food.

4 0
3 years ago
7. Sky Castle Company is conducting research on 3 independent communities of 12 members each to see how they have been consuming
tresset_1 [31]

Answer:

1.98% probability that at least 11 members are categorized as low risk participants in two of three communities

Step-by-step explanation:

To solve this question, we need to use two separate binomial trials.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Probability of at least 11 members being categorized as low risk participants.

12 members in the sample, so n = 12

30% chance to be categorized as high risk participants. So 100-30 = 70% probability of being categorized as low risk participants. So p = 0.7

This probability is

P(X \leq 11) = P(X = 11) + P(X = 12)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 1) = C_{12,11}.(0.7)^{11}.(0.3)^{1} = 0.0712

P(X = 2) = C_{12,12}.(0.7)^{12}.(0.3)^{0} = 0.0138

P(X \leq 11) = P(X = 11) + P(X = 12) = 0.0712 + 0.0138 = 0.0850

What is the probability that at least 11 members are categorized as low risk participants in two of three communities?

For each community, 8.50% probability of at least 11 members being categorized as low risk. So p = 0.085

Three comunities, so n = 3

This probability is P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{3,2}.(0.085)^{2}.(0.915)^{1} = 0.0198

1.98% probability that at least 11 members are categorized as low risk participants in two of three communities

6 0
4 years ago
PLEASE HELP 15 POINTS
yawa3891 [41]

Step-by-step explanation:

The sample space is the list of possible combinations:

H1, H2, H3, H4, T1, T2, T3, T4

Combinations that include a 2 are:

H2, T2

3 0
3 years ago
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