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2 years ago

Pamela is 6 years younger than Jiri. The sum of their ages is 78 . What is Jiri's age?

1 answer:

6 0

Let’s make

P = Pamela
J = Jiri

J - 6 = P

P + J = 78

We can replace P with J - 6 because they are the same thing.

J - 6 + J = 78

2J - 6 = 78

2J = 84

J = 42

Jiri is 42 years old.

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subtracting 1 to the product of 5 and x gives the same value as adding 10 to the product of 3 and x. what value of x makes the s

STatiana [176]

Answer:

true statement that makes

7 0

3 years ago

The sum of the arithmetic progression 4, ..., 76 is 1920. Find the number of terms and the common difference.

valkas [14]

Number of terms = 48

common difference = 1.5

This question involves the concept of Arithmetic Progression.

The formula for sum of an arithmetic progression series with first and last term given is;

$S_{n}$ = $\frac{n}{2}$ (a + l)

where;

a = first term

l = last term

n = number of terms

From the given sequence, we see that;

first term; a = 4

last term; l = 76

Sum of A.P; $S_{n}$ = 1920

Plugging in relevant values into the sum of an AP formula, we have;

1920 = $\frac{n}{2}$ (4 + 76)

simplifying this gives;

1920 = 40n

n = 1920/40

n = 48

Formula for nth term of an AP is;

$t_{n}$ = $a_{1}$ + (n - 1)d

where;

$a_{1}$ is first term

d is common difference

n is number of term

$t_{n}$ is the nth term in question

the 48th term is 76

Thus;

76 = 4 + (48 - 1)d

76 - 4 = 47d

72 = 47d

d = 72/47

d ≈ 1.5

Thus;

Number of terms = 48

common difference = 1.5

Read more at; brainly.com/question/16935540

4 0

3 years ago

A number is selected randomly from the integer from 10to30 what is the probability of picking a multiple of 4or5

arlik [135]

Answer:

$\frac{3}{7}$

Step-by-step explanation:

Start by finding how many integers there are from 10-30...

30-10+1=20+1=21

Note we have to add 1 since it is inclusive.

Now, let's find how many multiples of 4 or 5 there are from 10-30...

4*3=12

4*4=16

4*5=20

4*6=24

4*7=28

5*2=10

5*3=15

5*4=20

5*5=25

5*6=30

5+5-1=9

Note we have to subtract 1 since 20 is counted twice.

The probability would be...

$\frac{Multiples\ of\ four\ and\ five\ from\ 10-30}{Amount\ of\ integers\ between\ 10-30}=\frac{9}{21}=\frac{3}{7}$

6 0

2 years ago

48 ounces 6 cups ?? 2 cups HELP a.12 b.16 c.24 d.30

uranmaximum [27]

The answer is b 16 caz I got it on the website www.asknumbers.com

7 0

3 years ago

X/2 - y/3 = 3/2 X/3 + y/2 = 16/3

Ivenika [448]

X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
3x-2y=9______(1)

x/3+y/2=16/3
(6×x/3)+(6×y/2)=6×16/3
2x+3y=32_____(2)
(1)×3____9x-6y=27____(3)
(2)×2____4x+6y=64____(4)
(3)+(4)___13x=91
x=7
3(7)-2y=9
-2y=-12
y=6

4 0

2 years ago