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2 years ago

Tamika ran 120 meters in 21.6 seconds. Enrique ran 120 meters in 22.4

1 answer:

8 0

Tamika ran the fastest time at 21.6 seconds in 120 meters because the faster you run the less time you have on your timer.

You might be interested in

2.789 in expanded form

FromTheMoon [43]

It is...........

2.789=2 ones and 7 tenths 8 hundreths 9 thousandths

4 0

3 years ago

The railroad company wants to sell $8,134 in tickets. If each ticket costs $7, how many

Licemer1 [7]

Answer:

1162 tickets will need to be sold

Step-by-step explanation:

8134 / 7 = 1,162

8 0

2 years ago

Don put $4,000 in a savings account with an interest rate of 4% for three years. If the interest is compounded annually, how muc

miskamm [114]

Answer:

$4,499.46

Step-by-step explanation:

We can use the compound interest formula for this problem:

$A=P(1+\frac{r}{n} )^{nt}$

P = initial balance

r = interest rate (decimal)

n = number of times compounded annually

t = time

First, lets change 4% into a decimal:

4% -> $\frac{4}{100}$ -> 0.04

Now lets plug the values into the equation as shown below:

$A=4,000(1+\frac{0.04}{1})^{1(3)}$

$A=4,499.46$

Don will have $4,499.46 at the end of the three years.

4 0

2 years ago

2.516x-4)=104 (1:5+0.5x)

Andre45 [30]

Answer:

x= 42.92527821

Step-by-step explanation:

2.156x-4 = 104

x would be

42.92527821

_______________________________________________________

(1:5+0.5x)

and nothing farther can be done with this one

5 0

2 years ago

An automobile manufacturer claims that its car has a 35.935.9 miles/gallon (MPG) rating. An independent testing firm has been co

Serjik [45]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean.

For the given claim , we have

Null hypothesis : $H_0: \mu=35.9$

Alternative hypothesis : $H_a: \mu\neq35.9$

Since alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n=220 ;

Sample mean: $\overline{x}=35.6$ ;

Standard deviation: $s=2.2$

Test statistic for population mean:

$z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$z=\dfrac{35.6-35.9}{\dfrac{2.2}{\sqrt{220}}}\approx-2.02$

By using the standard normal distribution table of z , we have

P-value ( two tailed test ) : $2P(Z>|z|)=2(1-P(Z$

$=2(1-P(z$

Since , the P-value is greater than the significance level of $\alpha=0.02$ , it means we do not have sufficient evidence to reject the null hypothesis.

5 0

3 years ago