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3 years ago

7

Two identical blue boxes together weigh the same as three identical red boxes together. Each red box weighs 15.2 ounces. How muc

h does one blue box weigh in ounces?

2 answers:

Westkost [7]3 years ago

8 0

Answer:

Each blue box weighs 22.8 ounces

Step-by-step explanation:

dangina [55]3 years ago

8 0

Answer:

22.8 ounces

Step-by-step explanation:

Since one red box weighs 15.2 ounces, three red boxes weigh 45.6 ounces. This is the same as two blue boxes, so we have the equation 2b=45.6 where b represents the weight of the blue box. Solving the equation by multiplying both sides by 1/2 gives us 22.8 ounces.

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Steph,andra,emily,and becca are planning a road trip from their hometown to san antonio

morpeh [17]

Answer:

a. Emily should begin her turn as the third driver at point (1, -0.5).

b. Emily's turn to drive end at point (-2.5, -3.75).

Step-by-step explanation:

Let assume that the group of girls travels from their hometown to San Antonio in a straight line. We know that each location is, respectively:

Hometown

$H(x,y) = (8,6)$

San Antonio

$T(x,y) = (-6,-7)$

Then, we can determine the end of each girl's turn to drive by the following vectorial expression based on the vectorial equation of the line:

Steph

$S(x,y) = H(x,y) + \frac{1}{4}\cdot [T(x,y)-H(x,y)]$ (1)

$S(x,y) = (8,6) + \frac{1}{4}\cdot [(-6,-7)-(8,6)]$

$S(x,y) = (8,6) +\frac{1}{4}\cdot (-14,-13)$

$S(x,y) = (4.5,2.75)$

Andra

$A(x,y) = H(x,y) + \frac{2}{4}\cdot [T(x,y)-H(x,y)]$ (2)

$A(x,y) = (8,6) + \frac{2}{4}\cdot [(-6,-7)-(8,6)]$

$A(x,y) = (8,6)+\frac{2}{4}\cdot (-14,-13)$

$A(x,y) =(1, -0.5)$

Emily

$E(x,y) = H(x,y) + \frac{3}{4}\cdot [T(x,y)-H(x,y)]$ (3)

$E(x,y) = (8,6) + \frac{3}{4}\cdot [(-6,-7)-(8,6)]$

$E(x,y) = (8,6)+\frac{3}{4}\cdot (-14,-13)$

$E(x,y) = (-2.5, -3.75)$

a. <em>If the girls take turns driving and each girl drives the same distance, at what point should they stop from Emily to begin her turn as the third driver? </em>

Emily's beginning point is the Andra's stop point, that is, $A(x,y) =(1, -0.5)$ .

Emily should begin her turn as the third driver at point (1, -0.5).

b. <em>At what point does Emily's turn to drive end?</em>

Emily's turn to drive end at point (-2.5, -3.75).

7 0

3 years ago

How do I solve question 2b) Given that 16^m=4, write down the value of m.

mafiozo [28]

M=1/2=>√16=4
good luck.

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4 years ago

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Write your answer as a whole number and remainder.530/34

arsen [322]

Hey, Zalah. For my answer, I simplified 530/34. Which equals <span>15.58824. Hope this helps! :) </span>

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3 years ago

A conditional relative frequency table is generated by row from a set of data. The conditional relative frequencies of the two c

Gre4nikov [31]

The right answer for the question that is being asked and shown above is that: "D.) There is likely an association between the categorical variables because the relative frequencies are both close to 0.50." Given that a relative frequencies of 0.48 and 0.52, there will be an association between the categorical variables.<span>
</span>

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3 years ago

Read 2 more answers

According to government data, 75% of employed women have never been married. Rounding to 4 decimal places, if 15 employed women

blagie [28]

Answer:

We are given that According to government data, 75% of employed women have never been married.

So, Probability of success = 0.75

So, Probability of failure = 1-0.75 = 0.25

If 15 employed women are randomly selected:

a. What is the probability that exactly 2 of them have never been married?

We will use binomial

Formula : $P(X=r) =^nC_r p^r q^{n-r}$

At x = 2

$P(X=r) =^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=2) =\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{13}$

$P(X=2) =8.8009 \times 10^{-7}$

b. That at most 2 of them have never been married?

At most two means at x = 0 ,1 , 2

So, $P(X=r) =^{15}C_0 (0.75)^0 (0.25^{15-0}+^{15}C_1 (0.75)^1 (0.25^{15-1}+^{15}C_2 (0.75)^2 (0.25^{15-2}$

$P(X=r) =(0.75)^0 (0.25^{15-0}+15 (0.75)^1 (0.25^{15-1}+\frac{15!}{2!(15-2)!} (0.75)^2 (0.25^{15-2})$

$P(X=r) =9.9439 \times 10^{-6}$

c. That at least 13 of them have been married?

P(x=13)+P(x=14)+P(x=15)

$={15}C_{13}(0.75)^{13} (0.25^{15-13})+{15}C_{14} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=\frac{15!}{13!(15-13)!}(0.75)^{13} (0.25^{15-13})+\frac{15!}{14!(15-14)!} (0.75)^{14}(0.25^{15-14}+{15}C_{15} (0.75)^{15} (0.25^{15-15})$

$=0.2360$

8 0

3 years ago