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3 years ago

Sum of 3 by 4 and 7 by 4 is ___

Mathematics

1 answer:

denpristay [2]3 years ago

3 0

Step-by-step explanation:

Sum of 3 by 4 and 7 by 4 is

$= \frac{3}{4} + \frac{7}{4 } \\ = \frac{3 + 7}{4} \\ = \frac{10}{4} \\ = \frac{5}{2}$

Hope it will help :)

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when gail goes on a hike she always packs a ratio of 20 protien bars to 15 water bottles in her backpack . for a short hike ,gai

dimulka [17.4K]

Answer:

4 bars

Step-by-step explanation:

Gail's ratio of bars to bottles is ...

20 : 15 = 4 : 3

For 3 bottles, she should pack 4 bars.

8 0

4 years ago

A(x) = x2. What is g(x)?

Y_Kistochka [10]

Answer:

Step-by-step explanation:

5x² - 3 - (x² - 4x - 8)

5x² - 3 - x² + 4x + 8

4x² + 4x + 5

6 0

3 years ago

Location is known to affect the number, of a particular item, sold by an auto parts facility. Two different locations, A and B,

Mama L [17]

We have two samples, A and B, so we need to construct a 2 Samp T Int using this formula:

$\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$

In order to use t*, we need to check conditions for using a t-distribution first.

Random for both samples -- NOT STATED in the problem ∴ proceed with caution!
Independence for both samples: 130 < all items sold at Location A; 180 < all items sold at Location B -- we can reasonably assume this is true
Normality: CLT is not met; n < 30 for both locations A and B ∴ proceed with caution!

Since 2/3 conditions aren't met, we can still proceed with the problem but keep in mind that the results will not be as accurate until more data is collected or more information is given in the problem.

Solve for t*:

We need the tail area first.

$\displaystyle \frac{1-.9}{2}= .05$

Next we need the degree of freedom.

The degree of freedom can be found by subtracting the degree of freedom for A and B.

The general formula is df = n - 1.

df for A: 13 - 1 = 12
df for B: 18 - 1 = 17
df for A - B: |12 - 17| = 5

Use a calculator or a t-table to find the corresponding t-score for df = 5 and tail area = .05.

t* = -2.015

Now we can use the formula at the very top to construct a confidence interval for two sample means.

$\overline {x}_A=39$
$s_A=8$
$n_A=13$
$\overline {x}_B = 55$
$s_B=2$
$n_B=18$
$t^{*}=-2.015$

Substitute the variables into the formula: $\displaystyle \overline {x}_1 - \overline {x}_2 \ \pm \ t^{*} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} }$ .