All categories

3 years ago

Abiodun and taiwo share 735 naira between them so that abiodun get 95naira more than taiwo find how much money each gets

Mathematics

1 answer:

DerKrebs [107]3 years ago

5 0

Answer:

Abiodun gets - 415 naira

Taiwo gets - 320 naira

Step-by-step explanation:

Let us consider x and y as Abiodun and Taiwo,

Therefore , acccording to the question , equation formed will be -

$x+y=735$ naira ------ 1

and

$x=y+95$

$x-y=95$ ------2

From equation 1 and 2, we get

$2x=830$

$x=\frac{830}{2}$

$x=415$ naira (that is Abiodun share)

now , Calculating Taiwo share from equation 1

$x+y=735\\415+y=735$

$y=735-415$

y = 320 naira (that is Taiwo share )

Therefore , Share of Abiodun = 415 naira

Share of Taiwo = 320 naira

You might be interested in

The College Board SAT college entrance exam consists of three parts: math, writing and critical reading (The World Almanac 2012)

Wittaler [7]

Answer:

Yes, there is a difference between the population mean for the math scores and the population mean for the writing scores.

Test Statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ .

Step-by-step explanation:

We are provided with the sample data showing the math and writing scores for a sample of twelve students who took the SAT ;

Let A = Math Scores ,B = Writing Scores and D = difference between both

So, $\mu_A$ = Population mean for the math scores

$\mu_B$ = Population mean for the writing scores

Let $\mu_D$ = Difference between the population mean for the math scores and the population mean for the writing scores.

Null Hypothesis, $H_0$ : $\mu_A = \mu_B$ or $\mu_D$ = 0

Alternate Hypothesis, $H_1$ : $\mu_A \neq \mu_B$ or $\mu_D \neq$ 0

Hence, Test Statistics used here will be;

$\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$ where, Dbar = Bbar - Abar

$s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$

n = 12

Student Math scores (A) Writing scores (B) D = B - A

1 540 474 -66

2 432 380 -52

3 528 463 -65

4 574 612 38

5 448 420 -28

6 502 526 24

7 480 430 -50

8 499 459 -40

9 610 615 5

10 572 541 -31

11 390 335 -55

12 593 613 20

Now Dbar = Bbar - Abar = 489 - 514 = -25

Bbar = $\frac{\sum B_i}{n}$ = $\frac{474+380+463+612+420+526+430+459+615+541+335+613}{12}$ = 489

Abar = $\frac{\sum A_i}{n}$ = $\frac{540+432+528+574+448+502+480+499+610+572+390+593}{12}$ = 514

∑ $D_i^{2}$ = 22600 and $s_D$ = $\sqrt{\frac{\sum D_i^{2}-n*(Dbar)^{2}}{n-1}}$ = $\sqrt{\frac{22600 - 12*(-25)^{2} }{12-1} }$ = 37.05

So, Test statistics = $\frac{Dbar - \mu_D}{\frac{s_D}{\sqrt{n} } }$ follows $t_n_- _1$

= $\frac{-25 - 0}{\frac{37.05}{\sqrt{12} } }$ follows $t_1_1$ = -2.34

Now at 5% level of significance our t table is giving critical values of -2.201 and 2.201 for two tail test. Since our test statistics doesn't fall between these two values as it is less than -2.201 so we have sufficient evidence to reject null hypothesis as our test statistics fall in the rejection region .

Therefore, we conclude that there is a difference between the population mean for the math scores and the population mean for the writing scores.

8 0

3 years ago

A man looking at a photograph says brothers and sisters I have none but that man’s father is my father son who is in the photogr

Ainat [17]

The father say to his son, that’s his father in the photograph.

6 0

3 years ago

Morse code is a way to transmit text using a series of dots or dashes The Morse code for "ADD" is shown What fraction of the sha

salantis [7]

5 shapes are dots because ADD on Morse code is •- -•• -••

5 0