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3 years ago

A man is 32 years old and his son is 5 years old. How many years hence will the

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

3 0

Answer:

Step-by-step explanation:

the current ages:

father= 32 yrs old

son=5 yrs old

-------

4(5+x) = 32+x

20+4x = 32+x

4x-x = 32–20

3x =12

x = 12/3

x = 4 years

------------------------------------

after the 4 years later the farther age will be 36 ( 32+4 = 36)

the son will be 9 ( 5+4= 9 years)

the x presents how the father will be after 4 times the age of the son

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School x and y are in a comepetition. School x has 8 more points than school y and 3 times more points as school y . How many po

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Answer: School Y has 4 points and School X has 12 points.

Explanation:

Let the number of points School Y has be x

Let the number of points School X has be 3x

According to question, we get

$3x-x=8\\\\2x=8\\\\x=\frac{8}{2}\\\\x=4$

So, the number of point School Y has 4.

and number of points School X has

$3x=3\times 4=12$

Hence, School Y has 4 points and School X has 12 points.

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3 years ago

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3 years ago

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

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Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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3 years ago