All categories

3 years ago

A bubble may be enclosed in a square whose side

Mathematics

1 answer:

Alchen [17]3 years ago

6 0

Assume 1 small bub side = 2cm

area = 2 × 2

= 4 cm²

total area of small bub

4 × 4 = 16cm²

area of big bub = total area of small bub

= 16cm²

one side of big bub

16 ÷ 4 = 4

1 small bub side : 1 big bub side

2 : 4

1 : 2

the side length of the large square is 2 times bigger than the side length of one small square.

You might be interested in

What is the degree of the missing angle?

Gre4nikov [31]

Answer
180°-(49+90+x)
x=180°- 139
X=41°
Step-by-step explanation

3 0

2 years ago

D + 4/3 = -1/3<br> find d.

vovangra [49]

Answer:

d= -5/3

Step-by-step explanation:

d + 4/3 = -1/3

d= -4/3 -1/3

d= -5/3

4 0

3 years ago

Read 2 more answers

The temperature is -3.5 degrees Fahrenheit at 7:00 am. During the next 4 hours, the temperature decreases by -15.5 degrees. What

Kruka [31]

-19 degrees Fahrenheit

7 0

3 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.