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2 years ago

Estimate the percent using decimals 52% of 71

Mathematics

1 answer:

adelina 88 [10]2 years ago

4 0

1.732394366197183 so 2

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Damien bowled 5 games, His scores were:

Nat2105 [25]

The mean is basically the average.
To find the mean we need to add up all the values and divide it by the amount of games he played.
82 + 83 + 65 + 84 + 76 = 390
390/5
Mean = 78

7 0

3 years ago

Alexis wants to make a paperweight at pottery class. He designs a pyramid-like mode l with a base area of 100 square centimeters

olga55 [171]

Answer:

$\frac{3}{2}$ g/cm²

Step-by-step explanation:

First, we need to find the volume of the prism. The volume of a square pyramid is V = Bh/3

B = 100 cm²

h = 6 cm

V = 100 cm² * 6 cm /3

V = 600 cm³ /3

V = 200 cm³

Next, we need to solve for the density based on the desired volume and mass.

m = mass = 300 g

V = volume

density = m/V

density = 300 g / 200 cm³

density = 3/2 g/cm³

The lowest possible density of the material to make the paperweight would be $\frac{3}{2}$ g/cm²

4 0

2 years ago

1466899876+245689999999&&

tekilochka [14]

please mark this answer as brainlist

8 0

3 years ago

Pressure= 20N/10m2=

defon

Answer:

2PA

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Based on historical data, your manager believes that 41% of the company's orders come from first-time customers. A random sample

TEA [102]

Answer:

The probability that the sample proportion is between 0.35 and 0.5 is 0.7895

Step-by-step explanation:

To calculate the probability that the sample proportion is between 0.35 and 0.5 we need to know the z-scores of the sample proportions 0.35 and 0.5.

z-score of the sample proportion is calculated as

z= $\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of first time customers
p is the proportion of first time customers based on historical data

N is the sample size

For the sample proportion 0.35:

z(0.35)= $\frac{0,35-0.41}{\sqrt{\frac{0.41*0.59}{72} } }$ ≈ -1.035

For the sample proportion 0.5:

z(0.5)= $\frac{0,5-0.41}{\sqrt{\frac{0.41*0.59}{72} } }$ ≈ 1.553

The probabilities for z of being smaller than these z-scores are:

P(z<z(0.35))= 0.1503

P(z<z(0.5))= 0.9398

Then the probability that the sample proportion is between 0.35 and 0.5 is

P(z(0.35)<z<z(0.5))= 0.9398 - 0.1503 =0.7895

6 0

3 years ago