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jolli1 [7]
3 years ago
6

The average height of a girl in a middle school class is 54 inches. The average height of a boy is 55 inches. Find the ratio of

the average height of a girl to the
average height of a boy in this middle school class in the form a : b. Then write the ratio in fraction form.
Mathematics
1 answer:
kap26 [50]3 years ago
3 0

Answer:

1:1

Step-by-step explanation:

or 54:55

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4' 9''+ 7''<br> the sum pleaseee
joja [24]

Answer:

5'4" or 64 inches

Step-by-step explanation:

5 0
3 years ago
Out triangle-shaped pennants on the first day of school. If each pennant has a base of 5 inches and a height of 12 inches, what
brilliants [131]

Answer:

Area of the penny = 30 square inches

Step-by-step explanation:

Out triangle-shaped pennants on the first day of school. If each pennant has a base of 5 inches and a height of 12 inches, what is the area of each pennant in square inches?

The pennant is triangular in shape. Hence:

Area of a Triangle = 1/2× Base × Height

Base of the pennant = 5 inches

Height = 12 inches

Area of the pennant = 1/2 × 5 inches × 12 inches

= 30 square inches

3 0
3 years ago
Unit Exam: Rational Numbers
Minchanka [31]
Assuming that you’re asking for a fraction of 0.31 where 31 repeats?

31/99
8 0
3 years ago
Please help i know this is not that hard but i have this and one other thing due in a few minutes and i can't get them both in o
HACTEHA [7]

Answer:

0.2350 x 180 = 42.30

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Da
alexgriva [62]

Answer:

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>Recorded here are the germination times (in days) for ten randomly  chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.</em>

<em />

We start calculating the sample mean M and standard deviation s:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(18+12+20+17+14+15+13+11+21+17)\\\\\\M=\dfrac{158}{10}\\\\\\M=15.8\\\\\\

s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{\dfrac{101.6}{9}}\\\\\\s=\sqrt{11.3}=3.4\\\\\\

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{10}}=\dfrac{3.4}{3.162}=1.075

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=3.25 \cdot 1.075=3.49

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3

The 99% confidence interval for the mean germination time is (12.3, 19.3).

8 0
3 years ago
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