The distance 13,860 feet is equal to 2 5/8 write this as a decimal please

Mathematics

1 answer:

Ann [662]3 years ago

8 0

Answer: 2.625

Step-by-step explanation:

All you have to do is divide 5 by 8 which gives you .625 and then you put the 2 in front of the decimal amd you get your answer.

You might be interested in

Jane is making a suit which requires 2 5⁄8 yards for the jacket and 1 3⁄4 yards for the skirt. What's the total amount of materi

Ann [662]

The answer is B. 4 3⁄8 yards. The amount of material needed for the jacket is: x = 2 5/8 yd = 2 + 5/8 = 16/8 + 5/8 = (16+5)/8 = 21/8 yd. The amount of material needed for the skirt is: y = 1 3/4 yd = 1 + 3/4 = 4/4 + 3/4 = (4+3)/4 = 7/4 yd. The total amount of material Jane needs is: x + y = 21/8 + 7/4 = 21/8 + 14/8 = (21+14)/8 = 35/8 = (32+3)/8 = 32/8 + 3/8 = 4 + 3/8 = 4 3/8 yd.

7 0

4 years ago

The service life of a battery used in a cardiac pacemaker is assumed to be normally distributed. A random sample of ten batterie

Fittoniya [83]

Answer:

The confidence interval is $25.16 < \mu < 26.85$

Step-by-step explanation:

From the question we are given a data set

25.5, 26.1, 26.8, 23.2, 24.2, 28.4, 25.0, 27.8, 27.3, and 25.7.

The mean of the this sample data is

$\= x = \frac{\sum x_i}{n}$

where is the sample size with values n = 10

$\= x = \frac{25.5+ 26.1+ 26.8+23.2+ 24.2+ 28.4+ 25.0+ 27.8+ 27.3+ 25.7}{10}$

$\= x = 26$

The standard deviation is evaluated as

$\sigma = \sqrt{\frac{\sum (x-\= x)}{n} }$

substituting values

$= \sqrt{\frac{ ( 25.5-26)^2, (26.1-26)^2, (26.8-26)^2, (23.2-26)^2}{10} }$

$\cdot \ \cdot \ \cdot \sqrt{\frac{ ( 24.2-26)^2, (28.4-26)^2+( 25.0-26)^2+ (27.8-26)^2+( 27.3-26)^2+( 25.7-26)^2}{10} }$

$\sigma = 1.625$

The confidence level is given as 90% hence the level of significance is calculated as

$\alpha = 100 -90$

$\alpha =10$ %

$\alpha = 0.10$

Now the critical values of $\frac{\alpha }{2}$ is obtained from the normal distribution table as

$Z_{\frac{\alpha }{2} } = 1.645$

The reason we are obtaining the critical values of $\frac{\alpha }{2}$ instead of $\alpha$ is because we are considering two tails of the area under the normal curve