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STALIN [3.7K]
3 years ago
15

Round the weight 89 grams nearest kilogram​

Mathematics
1 answer:
tatiyna3 years ago
3 0

89 gram= 8.9 decagram = 0.89 hectograms = 0.089 kilogram

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A salesman's commission on 150,000 sale is 3000. what is the percent?
fomenos
The percentage is 98%.

You can find this by multiplying 98% (0.98) and 150,000 to get 147,000. Then subtract 147,000 from 150,000 to get 3,000.

I hope this helps!
8 0
4 years ago
Complete the following statement <br> (Picture above)
SOVA2 [1]
For this problem, you would replace x with 7 then solve.

3/ (7+2) - sqrt(7-3) =

3/9 - sqrt(4) =

1/3 - 2 = -1 2/3 = -1.67

f(7) = 1
6 0
3 years ago
Write the word sentence as a equation. Then solve the equation 13 subtracted from a number w is 15
patriot [66]
W-13=15 add 13 on both sides to isolate variable and you get w=13+15, w=28
7 0
3 years ago
Not sure. Please explain the steps.
stiv31 [10]

Answer:

answer C

Step-by-step explanation:

hello

because this is not allowed to divide by 0 I would take x real different from 0 and from -5/4

\dfrac{8x^3-10x^2-25x}{4x^2+5x}=\dfrac{x(8x^2-10x-25)}{x(4x+5)}=\\\dfrac{8x^2-10x-25}{4x+5}=\dfrac{(4x+5)(2x-5)}{4x+5}=2x-5

indeed if we develop

(4x+5)(2x-5)=8x^2-20x+10x-25=8x^2-10x-25

hope this helps

do not hesitate if you have any question

3 0
3 years ago
Read 2 more answers
Suppose 3% of the people contacted by phone are receptive to a certain sales pitch and buy your product. If your sales staff con
Shalnov [3]

Answer:

0% probability that more than 100 of the people contacted will purchase your product

Step-by-step explanation:

To solve this question, i am going to use the binomial approximation to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

p = 0.03, n = 2000

So

\mu = E(X) = np = 2000*0.03 = 60

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.03*0.97} = 7.63

What is the probability that more than 100 of the people contacted will purchase your product?

This is 1 subtracted by the pvalue of Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 60}{7.63}

Z = 5.24

Z = 5.24 has a pvalue of 1

1 - 1 = 0

0% probability that more than 100 of the people contacted will purchase your product

3 0
4 years ago
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