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BartSMP [9]
3 years ago
15

What is 0.95 rounded to the nearest tenth?

Mathematics
2 answers:
arlik [135]3 years ago
5 0
The answer is going to be 1.0
katen-ka-za [31]3 years ago
3 0
The answer would be 1.0.
You might be interested in
How many different committees can be formed from 6 teachers and 37 students if the committee consists of 4 teachers and 4 ​stude
Anna [14]

Answer:

The committee of 8 members can be selected in 990,675 different ways.

Step-by-step explanation:

The order in which the teachers and the students are chosen is not important, which means that the combinations formula is used to solve this question.

Combinations formula:

C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this question:

4 teachers from a set of 6.

4 students from a set of 37.

Then

T = C_{6,4}C_{37,4} = \frac{6!}{4!2!} \times \frac{37!}{4!33!} = 990675

The committee of 8 members can be selected in 990,675 different ways.

4 0
3 years ago
The length of a rectangle 8 more than the width .the area is 513 square inches .find the length and width of the rectangle
Pani-rosa [81]

Answer:

width =19 inches

length= 27 inches

Step-by-step explanation:

Let width be w inches so according to given scenario as length is 8 more than width so the equation becomes

l=8+w inches (Equation 1)

Area of Rectangle= L*w

Area= 513

L*w=513    (Equation 2)

Putting l's value from equation 1 in equation 2

(8+w)w=513

w^2+8w-513=0

Making Factors

w^2+27w-19w-513=0

w(w+27)-19(w+27)=0

(w-19)(w+27)=0

w=19, w=-27

as width cannot be negative so if we only consider w=19 then

l=8+19

l=27

3 0
3 years ago
One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r
Flura [38]

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

3 0
3 years ago
Read 2 more answers
You are choosing a song for your dance recital. Song A has 11 beats in 10 seconds. Song B has 7 beats in 6 seconds. Which song h
Pachacha [2.7K]
By greater if you mean faster, then you are comparing the fractions 11/10 and 7/6. Finding a common denominator, which is thirty, we get the fractions 33/30 and 35/30. We can see that 35>33, making Song B faster. 
8 0
3 years ago
A cone has a volume of 602.88 cm3 and a radius of 8 cm. What is the height of the cone
kotykmax [81]

Answer:

h≈9cm

Step-by-step explanation:

V=

π

r

2

h

3 h=

3

V

π

r

2

=

3

602.88

π

8

2

≈

8.99544cm

4 0
3 years ago
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