Ask question

Ask question

All categories

2 years ago

10

A salesman is carrying a big bag for selling a machine. In addition to the machine that the salesman wants to sell, there is som

e personal stuff. The weight of his machine is three kilogram and half a machine. In addition to the machine, there is a repair tool for the machine that weighs half kilogram and one-fourth of the machine. The personal stuff weighs four kilogram minus a fourth of the weight of the machine. Finally, the salesman has some supplies that weigh two-fifth of the machine and personal stuff combined. What is the total weight in the bag of salesman

1 answer:

stira [4]2 years ago

7 0

The total weight of the bag that the salesman is carrying for selling a machine including some personal stuff is 13.9 Kg.

<h3>What is a fraction?</h3>

A fraction is a rational number that consists of a numerator and a denominator, used to represent a part of the whole thing.

<h3>Calculation:</h3>

It is given that,

A salesman is carrying a big bag for selling a machine.

Consider the weight of the machine = w

So,

The weight of his machine is three kilograms and half a machine. I.e.,

w = 3kg +w/2

⇒ w/2 = 3kg

∴ w = 6kg

Thus, the weight of the machine is 6 Kg.

So, the weight of the repair tool for the machine = a half kilogram and one-fourth of the machine

⇒ 1/2Kg + 1/4w

⇒ 1/2 + 6/4

⇒ 2 Kg

The personal stuff weighs - four kilograms minus a fourth of the weight of the machine.

⇒ 4Kg - w/4

⇒ 4 - 6/4

⇒ 2.5 Kg

He has some supplies that weigh two-fifth of the machine and personal stuff combined.

⇒ 2/5(w + 2.5)

⇒ 2/5 × 8.5

⇒ 3.4 Kg

Thus, the total weight of the bag = 6 Kg + 2 Kg + 2.5 Kg + 3.4Kg = 13.9Kg.

Therefore, the total weight of the salesman's bag is 13.9 Kg.

Learn more about fractions here:

brainly.com/question/78672

#SPJ4

You might be interested in

Adding and subtracting With Scientific notation<br><br><br>Please help

Tom [10]

Answer:

0.131

Step-by-step explanation:

10^-2=0.01 so 0.01+6=0.06

0.06+0.071=0.131

4 0

4 years ago

A microwave manufacturing company has just switched to a new automated production system. Unfortunately, the new machinery has b

baherus [9]

Answer:

Null hypothesis: $\mu \leq 6$

Alternative hypothesis: $\mu > 6$

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$

$df=n-1=36-1=35$

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

$p_v =P(t_{(35)}>2)=0.0267$

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=6.5$ represent the mean time for the sample

$s=1.5$ represent the sample standard deviation for the sample

$n=36$ sample size

$\mu_o =6$ represent the value that we want to test

$\alpha=0.05,0.1$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 6 days, the system of hypothesis would be:

Null hypothesis: $\mu \leq 6$

Alternative hypothesis: $\mu > 6$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{6.5-6}{\frac{1.5}{\sqrt{36}}}=2$

Critical value and P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=36-1=35$

In order to calculate the critical value we need to find a quantile on the t distribution with 35 degrees of freedom that accumulates $\alpha$ on the right. Using the significance level of 0.05 we got:

$t_{crit}=1.690$ with the excel code:"=T.INV(0.95,35)"

And using the significance of 0.1 we got

$t_{crit}=1.306$ with the excel code:"=T.INV(0.90,35)"

Since is a one side right tailed test the p value would be:

$p_v =P(t_{(35)}>2)=0.0267$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05,0.1$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say that the true mean it's significantly higher than 6 at 5% and 10% of significance.

6 0

3 years ago

The function y = f (x) is graphed below. What is the average rate of change of the function f(x) on the interval - 1 ≤ x ≤ 3?

jonny [76]

Answer:

$\displaystyle -2$

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Algebra I</u>

Average Rate of Change: $\displaystyle \frac{f(b) - f(a)}{b - a}$

Step-by-step explanation:

<u>Step 1: Define</u>

Interval -1 ≤ x ≤ 3

a = -1, b = 3

f(a) = f(-1) = 4

f(b) = f(3) = -4

<u>Step 2: Find Average</u>

Substitute in variables [ARC]: $\displaystyle \frac{f(3) - f(-1)}{3 + 1}$
Substitute: $\displaystyle \frac{-4 - 4}{3 + 1}$
[Fraction] Subtract/Add: $\displaystyle \frac{-8}{4}$
[Fraction] Divide: $\displaystyle -2$

4 0

3 years ago

Please help me, just need the answer and thank you!

Triss [41]

Answer to what? I’m confused lol

6 0

4 years ago

Read 2 more answers

Help please. i am desperate

Nana76 [90]

Answer:

lol

Step-by-step explanation:

whyyy u say that hahahahhaa

4 0

3 years ago