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3 years ago

5.62 × 0.7 = 2.68 × 1.35 True or false?

Mathematics

2 answers:

In-s [12.5K]3 years ago

7 0

That is completely wrong, the answer is false

konstantin123 [22]3 years ago

3 0

False, I multiplied both and the answers do not match

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For $7.52, you purchased 8 pens and highlighters from a local bookstore. Each highlighter cost $1.09 and each pen cost $.69. How

seraphim [82]

i can. it says you got 8 pens. 1.09h + .69(8) = 7.52 1.09h + 5.52 = 7.52 do you know what to do from here?

5 0

2 years ago

4. Two cylinders have dimensions in the ratio of 2:3. The smaller cylinder has a surface area of 16 cm.

Zarrin [17]

Answer:

(cross multiply)

2:3

16:X

2X=16×3

2x=48

x=48÷2

x=24

8 0

3 years ago

____ [38]

Answer:

It is 25 units

Step-by-step explanation:

The triangles are equal

5 0

3 years ago

Read 2 more answers

Determine the area of the rectangular ABCD with AB=5cm and BC=3cm. A 5cm B 3 cm D C с Determine the area of the triangular AB wi

lbvjy [14]

Ac is 20cm and an is 5 cm u jus have to add them together

7 0

2 years ago

In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p

Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is $\frac{23}{30}$ or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ ---------------(1)

where

P(A) = Probability of choosing a student playing basket ball

P(B) = Probability of choosing a student playing base ball

P(A \cap B) = Probability of choosing a student playing both the games

Finding P(A)

P(A) = $\frac{\text { Number of students playing basket ball }}{\text{Total number of students}}$

P(A) = $\frac{19}{30}$ --------------------------(2)

Finding P(B)

P(B) = $\frac{\text { Number of students playing baseball }}{\text{Total number of students}}$

P(B) = $\frac{12}{30}$ ---------------------------(3)

Finding $P(A \cap B)$

P(A) = $\frac{\text { Number of students playing both games }}{\text{Total number of students}}$

P(A) = $\frac{8}{30}$ -----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

$P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{31}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{23}{30}$

7 0

3 years ago