What is 85.807 rounded to the nearest ten?

Mathematics

2 answers:

olga55 [171]4 years ago

4 0

85.8 is the answer to this question

Serggg [28]4 years ago

4 0

85.8
because 85.807 becomes 85.81 which becomes 85.8, the .8 is in the tenth place.

You might be interested in

A car travels at 65 miles per hours. Going through construction it travels at 3/5 this speed. Write this fraction as a decimal a

olga nikolaevna [1]

3/5 = 6/10 = 0.6
0.6×65=39
Car travels 39 miles per hour through construction.

6 0

3 years ago

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 46 o

meriva

Answer:

(a) 68% of the widget weights lie between 43 ounces and 49 ounces.

(b) The percentage of the widget weights lie between 43 and 87 ounces is 15.87%.

Step-by-step explanation:

Let X = weight of widgets manufactured by Acme Company.

The distribution of the random variable X is, N (μ = 46, σ² = 3²).

According to the Empirical Rule in a normal distribution with mean µ and standard deviation σ, nearly all the data will fall within 3 standard deviations of the mean. The empirical rule can be broken into three parts:

68% data falls within 1 standard deviation of the mean. That is P (µ - σ ≤ X ≤ µ + σ) = 0.68.
95% data falls within 2 standard deviations of the mean. That is P (µ - 2σ ≤ X ≤ µ + 2σ) = 0.95.
99.7% data falls within 3 standard deviations of the mean. That is P (µ - 3σ ≤ X ≤ µ + 3σ) = 0.997.

(a)

According to the Empirical rule, 68% data falls within 1 standard deviation of the mean.

P (µ - σ ≤ X ≤ µ + σ) = 0.68.

Compute the upper and lower values as follows:

µ - σ = 46 - 3 = 43 ounces

µ + σ = 46 + 3 = 49 ounces

Thus, 68% of the widget weights lie between 43 ounces and 49 ounces.

(b)

Compute the probability of the widget weights lie between 43 and 87 ounces as follows:

$P(43$