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3 years ago

Four out of nine dogs have spots. What is the decimal equivalent for the number of dogs who have spots?

Mathematics

1 answer:

pishuonlain [190]3 years ago

5 0

Answer: 0.4444444444
Infinite # of 4 so just put a pin over it like this
_
0.4

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Two quantities are related, as shown in the table:

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Answer:

$y = \frac{1}{2} x + 2$ .

Step-by-step explanation:

From the given table it is clear that x and y are linearly related.

Now, any two points that satisfy the relation between x and y are sufficient to express the equation.

The first two given points are (2,3), and (4,4).

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3 years ago

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4 years ago

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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

Can you show me the graph y=-2×5^x

rewona [7]

This is the graph for y=-2(times)5^x
Hope this helps!

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3 years ago

Four out of nine dogs have spots. What is the decimal equivalent for the number of dogs who have spots?￼

Four out of nine dogs have spots. What is the decimal equivalent for the number of dogs who have spots?