A dime has a radius of 8 1/2 millimeters. Find the circumference of a dime to the nearest tenth.

1 answer:

8 0

Circumference=2pir
r=8.5
c=2pi8.5
c=17pi
aprox pi to 3.14
c=17 times 3.14
c=53.38
round to nearest 10th
53.4

answe ris 53.4mm

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Ann and Betty together have $60. Ann has $9 more than twice Betty's amount. How much money does each have?

Ymorist [56]

First and mainly keep that 9 one side,
There are 2 people involved Ann and Betty, Then divide the TOTAL AMOUNT BY THE NUMBER OF PEOPLE.
· Then it will look like $60÷2= $30 per person.
In this case everyone has equal amount of money, but the question say Ann has $9 more than Betty's. so, Ann≈ $30 +$9= $39.
Betty≈ $30-$9= $21
Because loses money because Ann has more,
TOTAL=Ann+Betty
$60= $39+$21

7 0

2 years ago

Suppose a baker claims that the average bread height is more than 15cm. Several of this customers do not believe him. To persuad

laila [671]

Answer:

$17-2.262\frac{1.9}{\sqrt{10}}=15.641$

$17+2.262\frac{1.9}{\sqrt{10}}=18.359$

So on this case the 95% confidence interval would be given by (15.641;18.359)

And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm

Step-by-step explanation:

Previous concepts

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 means 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=17$ represent the sample mean

$\mu$ population mean (variable of interest)

s=1.9 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=10-1=9$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that $t_{\alpha/2}=2.262$