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3 years ago

How many times can 49 go into 410?

Mathematics

2 answers:

Vinvika [58]3 years ago

7 0

Answer: 8 times

49 times 8 = 392

Step-by-step explanation:

Oxana [17]3 years ago

3 0

Answer:

8 times evenly without going over

Step-by-step explanation:

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Explain when each method(graphing, substitution,linear combinations) is a good method to find a solution to a system of equation

PSYCHO15rus [73]

Graphing is a good method when you have small integers or fractions as your slope.
Substitution is a good method when you have two equations, and when one of the coefficient of the variables is one.
Linear combinations is a good method when you have the same coefficient for the same variable in 2 different equations.

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3 years ago

Read 2 more answers

PLEASE HELP ME ITS MY LAST QUESTION!!

shepuryov [24]

100 = 10^2

1000 = 10^3

10000 = 10^4

0.1 = 10^(-1)

0.01 = 10^(-2)

0.001 = 10^(-3)

Now,

100 × 0.1 = 10^2 × 10^(-1)

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3 years ago

How many different ways can you choose 5 books from a shelf containing 12 books? Type a numerical answer in the space provided.

yanalaym [24]

There are 792 different ways of choosing 5 books from a shelf containing 12 books. This problem can be solved using the combination formula of mathematics. The combination formula of mathematics is a number of way of selecting thing from a collection and the formula stated as n(n-1)...(n-k+1)/k(k-1)...1 (Calculation: 12*11*10*9*8/5*4*3*2*1 = 95040/120 = 792).

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4 years ago

Determine the largest open rectangle in the ty-plane that contains the point (t0, y0) and in which the hypotheses of theorem 2.2

djyliett [7]

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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