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aalyn [17]
3 years ago
5

the speed of two insects equal the same speed when rounded to the nearest whole number; dragonfly:6.974 ;horsefly:3.934 ;bumbleb

ee:2.861 ;honeybee:2.548; housefly:1.917
Mathematics
2 answers:
Goryan [66]3 years ago
8 0
Am presuming you are looking for the two insects.

In rounding up to nearest whole number, you look at the number before the decimal point, if the number after that is 5 or more, you add up one extra to the digit before the decimal point.

When you round up bumblebee 2.861 to the nearest whole number you have 3.

When you round up honeybee 2.548 to the nearest whole number you have 3.

The two insects are bumblebee and honeybee.

No other two are equal when rounded up to the nearest whole number.
Vinvika [58]3 years ago
6 0
Dragonfly= 7
horsefly= 4
bumblebee= 3
honeybee= 3
housefly= 2
Dragonfly is the fastest with a speed of 7.
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Answer:2 in.

Step-by-step explanation:

Given

Dimension of photo frame is 20\ in.\times 24\ in.

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Suppose x be the width of border

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A'=(20-2x)(24-2x)

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So,

\Rightarrow (20-2x)(24-2x)=320

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\Rightarrow 120-10x-12x+x^2=80

\Rightarrow x^2-22x+40=0

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3 years ago
Test the claim that the mean GPA of night students is larger than 2 at the .025 significance level. The null and alternative hyp
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Answer:

H_0: \, \mu = 2.

H_1:\, \mu > 2.

Test statistics: z \approx 2.582.

Critical value: z_{1 - 0.025} \approx 1.960.

Conclusion: reject the null hypothesis.

Step-by-step explanation:

The claim is that the mean \mu is greater than 2. This claim should be reflected in the alternative hypothesis:

H_1:\, \mu > 2.

The corresponding null hypothesis would be:

H_0:\, \mu = 2.

In this setup, the null hypothesis H_0:\, \mu = 2 suggests that \mu_0 = 2 should be the true population mean of GPA.

However, the alternative hypothesis H_1:\, \mu > 2 does not agree; this hypothesis suggests that the real population mean should be greater than \mu_0= 2.

One way to test this pair of hypotheses is to sample the population. Assume that the population mean is indeed \mu_0 = 2 (i.e., the null hypothesis is true.) How likely would the sample (sample mean \overline{X} = 2.02 with sample standard deviation s = 0.06) be observed in this hypothetical population?

Let \sigma denote the population standard deviation.

Given the large sample size n = 60, the population standard deviation should be approximately equal to that of the sample:

\sigma \approx s = 0.06.

Also because of the large sample size, the central limit theorem implies that Z= \displaystyle \frac{\overline{X} - \mu_0}{\sigma / \sqrt{n}} should be close to a standard normal random variable. Use a Z-test.

Given the observation of \overline{X} = 2.02 with sample standard deviation s = 0.06:

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Because the alternative hypothesis suggests that the population mean is greater than \mu_0 = 2, the null hypothesis should be rejected only if the sample mean is too big- not too small. Apply a one-sided right-tailed z-test. The question requested a significant level of 0.025. Therefore, the critical value z_{1 - 0.025} should ensure that P( Z > z_{1 - 0.025}) = 0.025.

Look up an inverse Z table. The z_{1 - 0.025} that meets this requirement is z_{1 - 0.025} \approx 1.960.

The z-value observed from the sample is z_\text{observed}\approx 2.582, which is greater than the critical value. In other words, the deviation of the sample from the mean in the null hypothesis is sufficient large, such that the null hypothesis needs to be rejected at this 0.025 confidence level in favor of the alternative hypothesis.

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