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3 years ago

a herd of 86 horses has 75 white and some black horses.what is the ratio of black horses to white horses?

Mathematics

2 answers:

Naddik [55]3 years ago

8 0

Answer:11 black horses : 75 white horses

Step-by-step explanation:

we know thw white horses and how many there are but we dont know how many black ones so u take the number of white horses and subtart it from the total and get the number of black horses

timurjin [86]3 years ago

8 0

Answer:

11 : 75

Step-by-step explanation:

So first we find the number of black horses: 86 - 75 = 11 black horses

So the ratio is 11 : 75

We write 11 first because that is the number of horses asked for first.

HOPE IT HELPED

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Read 2 more answers

Unlike most packaged food products, alcohol beverage container labels are not required to show calorie or nutrient content. An a

zhenek [66]

Answer:

We conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

Step-by-step explanation:

We are given that an article reported on a pilot study in which each of 58 individuals in a sample was asked to estimate the calorie content of a 12 oz can of beer known to contain 153 calories.

The resulting sample mean estimated calorie level was 193 and the sample standard deviation was 88.

Let $\mu$ = true average estimated calorie content in the population sampled.

So, Null Hypothesis, $H_0$ : $\mu$ $\leq$ 153 calories {means that the true average estimated calorie content in the population sampled does not exceeds the actual content}

Alternate Hypothesis, $H_A$ : $\mu$ > 153 calories {means that the true average estimated calorie content in the population sampled exceeds the actual content}

The test statistics that would be used here One-sample t test statistics as we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean estimated calorie level = 193 calories

s = sample standard deviation = 88

n = sample of individuals = 58

So, the test statistics = $\frac{193-153}{\frac{88}{\sqrt{58} } }$ ~ $t_5_7$

= 3.462

The value of t test statistics is 3.462.

Since, in the question we are not given the level of significance so we assume it to be 5%. Now, at 0.05 significance level the t table gives critical value of 1.6725 at 57 degree of freedom for right-tailed test.

Since our test statistic is more than the critical value of t as 3.462 > 1.6725, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the true average estimated calorie content in the population sampled exceeds the actual content.

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