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3 years ago

Ava has two frogs. This is 1/3 the number of frogs that Heather has. How many frogs does Heather have?

Mathematics

2 answers:

lesya692 [45]3 years ago

6 0

Answer:

2x3=6 and that the number of frogs heather has or you can do this Ava has 2 frogs and heather has 1/3 more then take the 2 and add it to the 1 and that makes three. then take both of the threes and add them together and you then have the same number of 6

Step-by-step explanation:

uranmaximum [27]3 years ago

5 0

Answer:

All in all, Ava has 6 frogs.

Step-by-step explanation (how i figured it out):

If 1/3 equals 2 frogs. Then times 2 by 3 which gives you six. I'm bad at explaining, but hoped this somewhat helped.

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The sides of a triangular ground are 5mt, 7mt and 8mt respectively. Find the cost of levelling the ground at the rate of Rs10. P

AveGali [126]

Assume that the height, base and the hypothesis of the triangular ground is 5mt, 7mt and 8mt respectively:

Area of the triangular ground
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3 years ago

1. A coin-operated drink machine was designed to discharge a mean of 7 ounces of coffee per cup. In a test of the machine, the d

Lesechka [4]

Answer:

No, at the 0.1 level of significance, there is not enough evidence to conclude that the true mean discharge, µ, differs from 7 ounces.

Step-by-step explanation:

We are given that a coin-operated drink machine was designed to discharge a mean of 7 ounces of coffee per cup. In a test of the machine, the discharge amounts in 13 randomly chosen cups of coffee from the machine were recorded.

The sample mean and the sample standard deviation were 7.08 ounces and 0.22 ounces, respectively.

Let $\mu$ = true mean discharge amount

SO, Null Hypothesis, $H_0$ : $\mu$ = 7 ounces {means that the true mean discharge, µ, does not differs from 7 ounces}

Alternate Hypothesis, $H_a$ : $\mu \neq$ 7 ounces {means that the true mean discharge, µ, differs from 7 ounces}

The test statistics that will be used here is One-sample t test statistics because 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 discharge amount = 7.08 ounces

s = sample standard deviation = 0.22 ounces

n = sample of coffee cups = 13

So, test statistics = $\frac{7.08-7 }{{\frac{0.22}{\sqrt{13} } } }$ ~ $t_1_2$

= 1.311

Now at 0.1 significance level, the t table gives critical values between -1.782 and 1.782 at 12 degree of freedom for two-tailed test. Since our test statistics lies within the range of critical values of t so we insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the the true mean discharge, µ, does not differs from 7 ounces.

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3 years ago

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3 years ago

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Vika [28.1K]

A)
3 : 1
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B)
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C)
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sveticcg [70]

Answer:

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Step-by-step explanation:

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