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2 years ago

Find the percentage of students that scored higher than 95.5. Round your answer to the nearest percent.

Mathematics

1 answer:

My name is Ann [436]2 years ago

6 0

Answer:

$\% Change = \frac{9}{64} *100 = 14.0625\% \approx 14\%$

So then we have approximately 14% of the values higher than 95.5

Step-by-step explanation:

For this case we assume that the info is given on the figure attached.

From this figure we have the following frequencies for each class

Class Frequency

___________________________

70.5-75.5 13

75.5-80.5 7

80.5-85.5 6

85.5-90.5 10

90.5-95.5 19

95.5-100.5 9

____________________________

Total 64

So we have a total of 64 values and we want to find the percentage of students that scored higher than 95.5 so we can use the formula of relative change and we got:

$\% Change = \frac{9}{64} *100 = 14.0625\% \approx 14\%$

So then we have approximately 14% of the values higher than 95.5

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1 year ago

In 26,059, in which place is the 2? tens thousands hundreds ten thousands

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2 years ago

2 2/3 of what number is 6 1/7

borishaifa [10]

$2\frac{2}{3}$ of $2\frac{17}{56}$ is $6\frac{1}{7}$

Solution:

Given $2\frac{2}{3}$ of what number is $6\frac{1}{7}$ .

Let us first convert the mixed fraction into improper fraction.

$$2\frac{2}{3}=\frac{(2\times3)+2}{3}=\frac{8}{3}$

$$6\frac{1}{7}=\frac{(6\times7)+1}{3}=\frac{43}{3}$

Now, let us take the unknown number be x.

$$2\frac{2}{3}\times x=6\frac{1}{7}$

$$\frac{8}{3}\times x=\frac{43}{7}$

Do the cross multiplication.

$$ x=\frac{43}{7}\times\frac{3}{8}$

$$ x=\frac{43\times3}{7\times8}$

$$ x=\frac{129}{56}$

Now, again change the improper fraction into mixed fraction.

$$ x=2\frac{17}{56}$

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2 years ago

Allen's hummingbird (Selasphorus sasin) has been studied by zoologist Bill Alther.† Suppose a small group of 10 Allen's hummingb

ruslelena [56]

Answer: provided in the explanation segment

Step-by-step explanation:

(a). from the question, we can see that since that б is known, we can use standard normal, z.

we are asked to find an 80% confidence interval for the average weights of Allen's hummingbirds in the study region. What is the margin of error?

⇒ 80% confidence interval for the average weight of Allen's hummingbirds is given thus;

x ± z * б / √m

which is

3.15 ± 1.28 * 0.32/√10

= 3.15 ± 0.1295 = 3.0205 or 3.2795

(b). normal distribution of weight (c) б is known

(c). option (a) and (e) are correct

(d). from the question, let sample size be given as S

this gives';

1.28 * 0.32/√S = 0.15

√S = (1.28 * 0.32) / 0.15 = 2.73

S = 7.4529

cheers i hope this helps

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2 years ago