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2 years ago

Need help plsss is 3/100 equal to 0.300 ???

Mathematics

2 answers:

zheka24 [161]2 years ago

7 0

Answer:

Yes. 3/100 = 0.3

Step-by-step explanation:

3/100 converted to decimal form equals 0.3.

Allushta [10]2 years ago

5 0

Answer:

Step-by-step explanation:

If you divide 100 by 3 you will end up with 0.03, not 0.3.

Hope this helps!

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Rewrite 4n^2+2n as a factorial.

Shkiper50 [21]

Answer:

4n²+2n as a factorial is given as:

$= (2n+1)!/(2n-1)!$

Step-by-step explanation:

We are given an expression which has to be converted into factorial form.

The expression is as follows:

$4n^{2} + 2n\\ = 2n(2n+1)\\ = (2n+1)2n\\$

Now we know that 2n+1 and 2n differs by '1' and the next smaller term is '2n-1'.

Hence, multiplying and dividing by '(2n-1)!'; we get:

$= ((2n+1)(2n)(2n-1)!)/(2n-1)!$

we know that x(x-1)(x-2)! = x!, so:

$= (2n+1)!/(2n-1)!$

4 0

3 years ago

Factor completely. x2+7x−18 Enter your answer in the box. 50 POINTS

Aliun [14]

The first step to solving this is to write 7x as a difference
x² + 9x - 2x - 18
next you need to factor out x from the expression
x × (x + 9) - 2x - 18
then you'll need to factor out -2 from the expression
x × (x + 9) - 2 (x + 9)
lastly,, you need to factor out x + 9 from the expression
(x + 9) × (x - 2)
this means that the correct answer to your question is (x + 9) × (x - 2).
let me know if you have any further questions
:)

4 0

3 years ago

Read 2 more answers

Plz help ASAP you will get 10 points if you answer this correctly. It is Saying to fix joe’s calculations.

zimovet [89]

Answer:

Ok how can I help

Step-by-step explanation:

3 0

4 years ago

Alex claims that the product of x^2 and x^4+x^2-1 will be x^8+x^4-x^2. explain the error made by alex

Alekssandra [29.7K]

Answer:

Step-by-step explanation:

the product is a result of multiplication....so lets multiply

x^2 (x^4 + x^2 - 1) =

x^6 + x^4 - x^2

ok...his mistakes are occurring because he is multiplying the exponents when he is supposed to be adding them.

exponent rule : x^a * x^b = x^(a + b)

5 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)} {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago