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2 years ago

Help! I PUT 50 POINTS!

Mathematics

1 answer:

artcher [175]2 years ago

4 0

Answer:

i-4 1/8

ii- 2 3/100

iii- 10 123/1000

iv- 292/625

Step-by-step explanation:

...

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Can I get some help on this question please

mr Goodwill [35]

things were not as complicated when I was in middle school.

3 0

2 years ago

Sarah is paid biweekly. Her annual salary is $33,000.

bezimeni [28]

Answer:

$1269.23

Step-by-step explanation:

Since Sarah is paid biweekly (every 2 weeks), and there are 52 weeks in a year...

$52$ ÷ $2=26$

Sarah is being paid 26 weeks out of the year.

Divide 33,000 by 26 (I only list 4 places after the decimal):

$33000$ ÷ $26=1269.2307$

Round 1269.2307 to the nearest cent (hundredth):

$1269.23$

3 0

2 years ago

Round 5,600,679 to the nearest thousand

vampirchik [111]

The answer is 5,601,000

4 0

3 years ago

there are approximately 4021/4 meters around a typical running track sandra has challenged herself to run 10 maps a day for 5 da

Lostsunrise [7]

Total distance of a typical running track = $402 \frac{1}{4} meters$ .

Let us convert it into impoerper fraction first, we get

$\frac{402\times4+1}{4}=\frac{1609}{4}meters.$

Total number of maps in a day = 10 maps.

Number of days of running = 5 days.

Total number of maps of 5 days = 10 × 5 = 50 maps.

Total number of meters in 50 maps = $50\times\frac{1609}{4}=\frac{80450}{4}=20112\frac{1}{2}\ meters.$ .

<h2>Therefore, $20112\frac{1}{2}\ meters$ will sandra run is she meets her challenge</h2>

5 0

3 years ago

New legislation passed in 2017 by the U.S. Congress changed tax laws that affect how many people file their taxes in 2018 and be

stiks02 [169]

Answer:

a. Yes

b. 0.35

c. 0.5

E(x)=34.25

V(x)=73.1875

SD(x)=8.555

Step-by-step explanation:

x 20 25 30 35 40 45 50

f(x) 0.05 0.2 0.25 0.15 0.15 0.1 0.1

This will be valid probability distribution if

1. All probabilities are between 0 and 1 inclusive.

2. Sum of probabilities must be 1.

Condition 1 is satisfied in this distribution. Now, we will check second condition

sum(f(x))=0.05+0.2+0.25+0.15+0.15+0.1+0.1=1.

As the both conditions are satisfied, thus the given distribution is a valid probability distribution.

P(x≥40)=P(X=40)+P(X=45)+P(X=50)

P(x≥40)=0.15+0.1+0.1

P(x≥40)=0.35

So, the probability that Backens and Hayes LLC will obtain 40 or more new clients is 0.35.

P(x<35)=P(X=20)+P(X=25)+P(X=30)

P(x<35)=0.05+0.2+0.25

P(x<35)=0.5

So, the probability that Backens and Hayes LLC will obtain fewer than 35 new clients is 0.5.

Expected value of x=E(x)= sum[x*f(x)]

E(x)= 20*0.05+25*0.2+30*0.25+35*0.15+40*0.15+45*0.1+50*0.1

E(x)=1+5+7.5+5.25+6+4.5+5

E(x)=34.25

Variance of x=V(x)=sum[x²*f(x)]-(sum[x*f(x)])²

sum[x²*f(x)]= 20²*0.05 +25²*0.2 +30²*0.25 +35²*0.15 +40²*0.15 +45²*0.1 +50²*0.1

sum[x²*f(x)]=20+1255+225+183.75+240+202.5+50

sum[x²*f(x)]=1246.25

V(x)=1246.25-(34.25)²

V(x)=73.1875

Standard deviation of x=SD(x)=√V(x)

SD(x)=8.555

3 0

2 years ago