All categories

3 years ago

What is the average speed of a toy car that travels 240 cm in one minute?

Mathematics

2 answers:

pshichka [43]3 years ago

8 0

Answer:

A is correct

Step-by-step explanation:

average speed : distance / time = 240cm / 60sec = 4cm / sec : 4 centimeters per second

umka21 [38]3 years ago

3 0

Answer:

Step-by-step explanation:

To find the cars rate per second we need to divide 240 by 60

240/60 = 4

You might be interested in

Eli took the fleet of 8 vans for oil changes. All of the vans needed windshield wipers which cost $24 per van. The total bill wa

Luba_88 [7]

High School Mathematics 5+3 pts Eli took the fleet of 8 vans for oil changes. All of the vans needed windshield wipers which cost $24 per van. The total bill was $432 . Write an equation to find out what each oil change cost . Solve the equation. 8×24=192-432=240÷8=$30

3 0

3 years ago

Horatio transferred a balance of $2600 to a new credit card at the beginning

4vir4ik [10]

Answer:

Step-by-step explanation:

2600(1+0.043/12)^6(1+0.137/12)^7

8 0

3 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