Ask question

Ask question

All categories

3 years ago

10

Round each mixed number to the nearest whole number. Then, estimate the product.

2 answers:

Naddik [55]3 years ago

7 0

8 9/10 = 8.9

9 is greater than 5, so round up

8.9 becomes 9

1 4/5 = 1.8

8 is greater than 5, so round up

1.8 becomes 2

9 x 2 = 18

18 should be your answer

hope this helps

Agata [3.3K]3 years ago

3 0

Hi there!

First, let's round 8 9/10:

Since 9/10 is about 1, 8 9/10 is about 9

Now, let's round 1 4/5:

Since 4/5 is about 1 as well, 1 4/5 is about 2

9 * 2 = 18

So, the approximate answer is 18.

Hope this helps!

You might be interested in

Solve the inequality. Graph the solution.<br><br> w+6≤−3

yuradex [85]

Answer: w<-9 With the line underneath

Step-by-step explanation:

4 0

3 years ago

⅔ parts of the earth surface is covered with water.

Nat2105 [25]

Answer:

i.) 66% of the earth is covered in water

ii.) 33% is continental

Step-by-step explanation:

4 0

3 years ago

To solve -8p = 48, which of the following could you do to both sides of the equation? A. add -8 B. subtract -8 C. multiply by -8

masya89 [10]

Answer:

D. divide by -8

Step-by-step explanation:

8 0

3 years ago

Louis made 24 quarts of lemonade he divided the lemonade equally between 6 pitchers how many points of lemonade did Louis put in

Mandarinka [93]

Answer: Each pitcher has 4 quarts of lemonade.

Step-by-step explanation:

Louis made 24 quarts of lemonade, and he divided it equally into 6 pitchers.

Then, each pitcher must has a sixt of the total amount of lemonade, this is:

p = (1/6)*24 quarts= (24/6) quarts = 4 quarts

Each pitcher has 4 quarts of lemonade.

7 0

3 years ago

Read 2 more answers

A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normall

Vlad1618 [11]

Answer: (2.54,6.86)

Step-by-step explanation:

Given : A random sample of 10 parking meters in a beach community showed the following incomes for a day.

We assume the incomes are normally distributed.

Mean income : $\mu=\dfrac{\sum^{10}_{i=1}x_i}{n}=\dfrac{47}{10}=4.7$

Standard deviation : $\sigma=\sqrt{\dfrac{\sum^{10}_{i=1}{(x_i-\mu)^2}}{n}}$

$=\sqrt{\dfrac{(1.1)^2+(0.2)^2+(1.9)^2+(1.6)^2+(2.1)^2+(0.5)^2+(2.05)^2+(0.45)^2+(3.3)^2+(1.7)^2}{10}}$

$=\dfrac{30.265}{10}=3.0265$

The confidence interval for the population mean (for sample size <30) is given by :-

$\mu\ \pm t_{n-1, \alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given significance level : $\alpha=1-0.95=0.05$

Critical value : $t_{n-1,\alpha/2}=t_{9,0.025}=2.262$

We assume that the population is normally distributed.

Now, the 95% confidence interval for the true mean will be :-

$4.7\ \pm\ 2.262\times\dfrac{3.0265}{\sqrt{10}} \\\\\approx4.7\pm2.16=(4.7-2.16\ ,\ 4.7+2.16)=(2.54,\ 6.86)$

Hence, 95% confidence interval for the true mean= (2.54,6.86)

7 0

3 years ago