All categories

3 years ago

If the teacher has 30 students and 10% left,how much students does she have now

Mathematics

2 answers:

garri49 [273]3 years ago

8 0

Answer:

She has 27 students now.

Step-by-step explanation:

Given :

The teacher has 30 students

10% students left.

To find : how much students does she have now?

Solution :

Since we are given that teacher has 30 students .

And 10% of students left

⇒ 10% of 30

⇒ $\frac{10}{100} *30$

⇒ $\frac{300}{100}$

⇒ $3$

Thus, 3 students left .

She had 30 students . three students left.

So, now she have 30 - 3 = 27 students .

Hence she has 27 students now.

Juli2301 [7.4K]3 years ago

7 0

Step-by-step explanation:

Teacher has 30 students

10%decrease

So,decrease=10/100*30=3

So, decrease is 3 students

Hence remaining number of students = 30-3=27 students

You might be interested in

The oil tank in your car is leaking at a rate of 1.2 oz per mile driven. You drove 15 miles over to your grandma's house. How ma

belka [17]

2.25 ounces were leaked.

8 0

3 years ago

Read 2 more answers

T(z)= sin^2(z/4), {-3 ≤ z ≤ 8}

Usimov [2.4K]

Answer:

T´(z)=n^2s/4*i

Step-by-step explanation:

T(z)=d/dz(sin^2*z/4)

T´(z)=d/dz(sn^2zi/4)

T´(z)=d/dz(sn^2z/4*i)

T´(z)i*d/dz(sn^2z/4)

T´(z)=n^2s/4*i

Simplify

3 0

2 years ago

Help with algebra please

Lena [83]

Domain - [-3,3] - What x values there are
Range- [-3,1] - What y values there are

6 0

2 years ago

Read 2 more answers

Make an approximate conversion from English to metric. 9 ft to meters

Anna71 [15]

Answer:

2.7 meters

Step-by-step explanation:

I hope this helped

4 0

2 years ago

Computer keyboard failures are due to faulty electrical connects (12%) or mechanical defects (88%). Mechanical defects are relat

anzhelika [568]

Answer:

(c) Probability that a failure is due to loose keys = 0.2376

(d) Probability that a failure is due to improperly connected or poorly welded wires = 0.078

Step-by-step explanation:

The Whole probability scenario is given for Computer Keyboard failures.

(a) Let F be the event of failure due to faulty electrical connects, P(F) = 0.12

M be the event of failure due to mechanical defects, P(M) = 0.88

LK be the event of mechanical defect due to loose keys, P(LK/M) = 0.27

IA be the event of mechanical defect due to improper assembly, P(IA/M) =0.73

DW be the event of electrical connects due to defective wires,P(DW/F) = 0.35

IC be the event of electrical connects due to improper connections,

P(IC/F) = 0.13 .

PWW be the event of electrical connects due to poorly welded wires,

P(PWW/F) = 0.52

(b) Keyboard failures

Faulty electrical connects Mechanical Defects

P(F) = 0.12 P(M) = 0.88

Defective wires Improper Poorly Loose Keys Improper

P(DW/F)=0.35 Connections Welded wires P(LK/M)=0.27 Assembly

P(IC/F)=0.13 P(PWW/F)=0.52 P(IA/M)=0.73

This is the required tree diagram.

P(LK) =P(LK/M) * P(M) {This means mechanical failure is due to loose

keys}

P(LK) = 0.27 * 0.88 = 0.2376 .

(d) Probability that a failure is due to improperly connected or poorly welded

wires is given by P(IC $\bigcup$ PWW) ;

P(IC $\bigcup$ PWW) = P(IC) + P(PWW) - P(IC $\bigcap$ PWW) { Here P(IC $\bigcap$ PWW) = 0 }

P(IC) = P(IC/F) * P(F) = 0.13 * 0.12 = 0.0156

P(PWW) = P(PWW/F) * P(F) = 0.52 * 0.13 = 0.0676

Therefore, P(IC $\bigcup$ PWW) = 0.0156 + 0.0676 - 0 = 0.078 .

8 0

3 years ago