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3 years ago

Plz plz help it’s due at 12 Btw only number 12

Mathematics

1 answer:

Marta_Voda [28]3 years ago

4 0

Answer:

12.

Step-by-step explanation:

I'm sorry. I really don't know what you mean. I just wrote 12 because you want only the number twelve?

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Original price:$50;Markdown:22%;Retail price<br> Please show work

Komok [63]

22%=.20
50*.20=11
mark=(+) number (adding)
50+11=61
ans:61
hope it help u:)

5 0

4 years ago

Find the ratio of : 800 ml to 4.8 litres

Ray Of Light [21]

Answer:

The ratio is $\frac{1}{6}$

Step-by-step explanation:

We have

$\frac{800mL}{4.8L} \\\\\frac{0.8L}{4.8L}\\\\\frac{1}{6}$

5 0

2 years ago

PLEASE ANSWER

Irina18 [472]

Answer:

You can by 8 of each item and you will still have money left over

Step-by-step explanation:

small=36

medium=52

large=52

that makes 134

6 0

3 years ago

Pls answer. i will mark u as branliest

kumpel [21]

Answer:

Step-by-step explanation:

All angles = 180

All sides are equal

Therefore, 180/3 = angle degrees = 60

x is an opposite angle of bottom right angle therfore x = 60

5 0

3 years ago

Read 2 more answers

Stress at work: In a poll conducted by the General Social Survey, 80% of respondents said that their jobs were sometimes or alwa

Svet_ta [14]

Answer:

a) P=0.019

b) P=0.263

c) P=0.794

Step-by-step explanation:

We assume that the poll gives the population's proportion of respondents said that their jobs were sometimes or always stressful (p=0.8).

Then, a sample of size n=190 is taken.

The sample mean is:

$\mu=np=190*0.8=152$

The sample standard deviation is:

$\sigma=\sqrt{np(1-p)}=\sqrt{190*0.8*0.2}=\sqrt{30.4}=5.514$

The probability that 140 or fewer workers find their jobs stressful is:

$z=(X-\mu)/\sigma=(140.5-152)/5.514=-11.5/5.514=-2.08 \\\\P(X\leq140)=P(X$

Note: a correction for continuity is applied.

The probability that more than 155 workers find their jobs stressful is

$z=(X-\mu)/\sigma=(155.5-152)/5.514=3.5/5.514= 0.635 \\\\P(X>155)=P(X>155.5)=P(z>0.635)=0.263$

The probability that the number of workers who find their jobs stressful is between 145 and 158 inclusive is:

$z_1=(X_1-\mu)/\sigma=(158.5-152)/5.514=6.5/5.514= 1.179\\\\ z_2=(X_2-\mu)/\sigma=(144.5-152)/5.514=-7.5/5.514= -1.360 \\\\ P(145\leq X\leq 158)=P(144.5< X< 158.5)\\\\P(145\leq X\leq 158)=P(X$

5 0

3 years ago

Read 2 more answers