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4 years ago

7

What is 600000 minus 145782

2 answers:

Vlad1618 [11]4 years ago

8 0

The answer is 454218 I checked on a calculator lol

svp [43]4 years ago

4 0

454,218 Is The Answer

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30 points.. Please Help! I don't understand what to do with this question...a bit confused.

Romashka-Z-Leto [24]

Eh, here's my best try for a table:

<u>Like tv</u> <u> don't like tv</u> <u>total</u>
<u>Like reading:</u> 70 10 80
<u>Don't like read</u>. <u>20</u> <u>30</u> <u>50 </u>
<u>total </u> 90 40 130

The percentage of students that like tv as well as reading is- 70 divided by 130 which gives you .538 which in percentage is 53.8%.

The probability of a student that does not prefer to watch tv as also does not like to spend their time reading is 30 over 40 which can be simplified or reduced to 3 over 4. [30/40 reduced to 3/4]

Hope this somehow helped you :)

5 0

3 years ago

Explain how multiplying fractions is similar to mulitiplying whole numbers and how it is different

tigry1 [53]

You could find the difference on the internet if not found tell me and I would explain for you

3 0

3 years ago

Read 2 more answers

What’s the value of x ?

scoray [572]

Answer: 10

Hope this helps you :)

6 0

2 years ago

Can you please help me with this I’ll give you a playlist for sure it’s

lapo4ka [179]

Answer:

15+(-37)

initial 15

fell 37 so -37

7 0

3 years ago

Read 2 more answers

Suppose a certain study reported that 27.7% of high school students smoke.

Norma-Jean [14]

<u>Given</u>:

Probability of student smoke,

P = 27.7%

= 0.277

Number of students,

n = 632

$q = 1-p$

$=1-0.277$

$=0.723$

(i)

Here,

Number of students (n) = 60

then,

⇒ $n_P=60\times 0.277$

$=16.62$

⇒ $n_q=60\times 0.723$

$=43.38$

We can see that $n_P > 10$ and $n_q>10$ so the normal approximation condition are met.

Now,

$\mu = n_P= 16.62$

$\sigma = \sqrt{n_{Pq}}$

$= \sqrt{60\times 0.277\times 0.723}$

$=3.9664$

Now,

⇒ $P(X$

$=P(Z_{18.5})$

The Z-score is:

= $\frac{18.5-16.62}{3.4664}$

= $0.5423$

hence,

The probability will be:

⇒ $P(Z_{18.5}) = 0.7062$

or,

⇒ $P(Z$

(ii)

Here,

Number of students (n) = 75

$\mu = n_P = 75\times 0.277$

$=20.775$

$\sigma = \sqrt{n_{Pq}}$

$=\sqrt{75\times 0.277\times 0.723}$

$=3.8756$

Now,

⇒ $P(X>17) = P(X> 17.5)$

$=1-P(X \leq 17.5)$

$=1-P(Z_{17.5})$

The Z-score is:

= $\frac{17.5-20.775}{3.8756}$

= $-0.9740$

then, $P(Z_{17.5}) = 0.165$

hence,

The probability will be:

⇒ $P(X>17) = 1-0.165$

$=0.835$

Learn more about probability here:

brainly.com/question/9825651

4 0

3 years ago