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2 years ago

an animal shelter has a ratio of dogs to cats test 3:2 if they are 30 cats at the shelter how many dogs are there

2 answers:

8 0

This ratio is in a form of dogs:cats.
The ratio given means there are 3 dogs for every 2 cats.
You can rewrite it as 3 dogs:2 cats.
If there are 30 cats, you would write the ratio as x dogs:30 cats.
Since the same must be done to both sides, you must find how many times 2 was multiplied to get 30 then multiply dogs by that number.
30 / 2 = 15
3 • 15 = 45
So if a shelter has 30 cats, then it has 45 dogs.
This can be written in a ratio as 45:30, or 45 dogs:30 cats.

lianna [129]2 years ago

5 0

45% are dogs in the shelter

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Approximately 80,000 marriages took place in the state of New York last year. Estimate the probability that for at least one of

sasho [114]

Answer:

a)0,45119

b)1

Step-by-step explanation:

For part A of the problem we must first find the probability that both people in the couple have the same birthday (April 30)

$P=\frac{1}{365} *\frac{1}{365}=\frac{1}{133225} \\$

Now the poisson approximation is used

λ=nP=80000*1/133225=0,6

Now, let X be the number of couples that birth April 30

P(X ≥ 1) =

1 − P(X = 0) =

$1-\frac{(e^-0.6)*(-0,6)^{0} }{0!}$

P(X ≥ 1) = 0,45119

B) Now want to find the

probability that both partners celebrated their birthday on th, assuming that the year is 52 weeks and therefore 52 thursday

$P=52*\frac{1}{365} *\frac{1}{365}=\frac{52}{133225} \\$

Now the poisson approximation is used

λ=nP=80000*52/133225=31.225

Now, let X be the number of couples that birth same day

P(X ≥ 1) =

1 − P(X = 0) =

$1-\frac{(e^-31.225)*(-31.225)^{0} }{0!}$

P(X ≥ 1) = 1

6 0

2 years ago

An experiment was performed to compare the wear of two different laminated materials. Twelve pieces of material 1 were tested by

GenaCL600 [577]

Answer:

At 0.05 level of significance, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units

Step-by-step explanation:

We hypothesize that mean difference between abrasive wear of material 1 and material 2 is greater than 2.

So we write the null hypothesis $H_0 : \mu_1 - \mu_2 >2$ ,

and the alternative hypothesis $H_1: \mu_1 - \mu_2 \leq 2$ .

We will find the T-score as well as the p-value. If the p-value is less than the level of significance, we will reject the null hypothesis, i.e. we will conclude that the abrasive wear of material 1 is less than that of material 2. Otherwise, we will accept the null hypothesis.

Since the variance is unknown and assumed to be equal, we will use the pooled variance

$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2} = 20.05$ ,

where $n_1 = 12, n_2 =10, s_1 =4, s_2 = 5$ .

The mean of material 1 and material 2 are $\mu_1 =85, \mu_2=81$ respectively and mean difference $d$ is equal to 4. The hypothesize difference $d_0$ is equal to 2.

To find the T-score, we use the following formula

$T = \frac{d - d_0}{\sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_2} }}$

Substituting all the values into the T-score formula gives us $T = 1.04$ , and the respective p-value is equal to 0.31. This means we have enough statistical evidence not to reject the null hypothesis, and at 5% significance level, the abrasive wear of material 1 exceeds that of material 2 by more than 2 units.

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2 years ago

Decrease £19064.67 by 9.5%<br>Give your answer rounded to 2 DP.<br>

Advocard [28]

Answer:

£17253.53 to the nearest hundredth.

Step-by-step explanation:

Decreasing by 9,5% is equivalent to multiplying by 1 - 0.095 = 0.905.

19064.67 * 0.905

= £17253.52635.

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storchak [24]

The nth term is 6n^2-11n+16

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2 years ago

Alberto is making a budget. He owes a cell phone bill of $60 per month for 23 more months, plus the fees for any extra usage suc

netineya [11]

I'm not sure what your book is saying about the matter but a cell phone is not usually an essential expense. If it is saying it is though, it would be an essential flexible expense.

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2 years ago

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