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alexdok [17]
2 years ago
14

The weight of Jacob’s backpack is made up of the weight of the contents of the backpack as well as the weight of the backpack it

self. Seventy percent of the total weight is textbooks. His notebooks weigh a total of 4 pounds, and the backpack itself weighs 2 pounds. If the backpack contains only textbooks and notebooks, which equation can be used to determine t, the weight of the textbooks?
Mathematics
2 answers:
Dafna11 [192]2 years ago
7 0
If the backpack (b) contains only textbooks (t) and notebooks (n), the total weight of the backpack is:

b+t+n

but notebooks are 4 pounds, so actually:

b+t+4

also, the backpack weights 2 pounds, so the total weight is:

2+t+4 =t+6
 now, "Seventy percent of <span> the total weight is textbooks. ", this means that


t=70%(t+6) - this equation can be used to determine the weight of the textbooks!
</span> 




puteri [66]2 years ago
3 0

Answer:

d. 0.7(t + 4 + 2) = t

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The prices of three t-shirts styles are $24, $30 and $36. the probability of choosing a $24 t-shirt is 1/6. the probability of c
Slav-nsk [51]

\text{Answer} : \text{The expected value of a t-shirt is \$31.}

Explanation:

Since we have given that

The prices of three t-shirts styles  i.e $24, $30, $36 with their probability is given by

\frac{1}{6}, \frac{1}{2},\frac{1}{3}

As we know that,

E(X)= \sum_{1}^{3}x_iP(x_i)

\text{where} x_i \text{ is the prices of t- shirts styles}

Now,

x_1= \$24 , x_2=\$30 , x_3=$36

and

P(x_1)=\frac{1}{6},P(x_2)=\frac{1}{2}, P(x_3)=\frac{1}{3}

So,

E(X)= 24\times \frac{1}{6}+30\times\frac{1}{2}+36\times \frac{1}{3}\\=4+15+12\\=31

So, the expected value of a t-shirt = $31.

4 0
2 years ago
A band director uses a coordinate plane to plan as how for a football game. During a show, the drummers will march along the lin
LenaWriter [7]

Answer:

The equation in slope-intercept form for the path of the trumpet players is y=\frac{x}{5}+\frac{12}{5}.

Step-by-step explanation:

Consider the provided information.

The drummers will march along the line y=-5x-8.

The trumpets players will march along a perpendicular line that passes through (-2,2).

The slope of line y=-5x-8 is: m₁ = -5.

The slope of perpendicular lines are: m_1\times m_2=-1

-5\times m_2=-1

m_2=\frac{1}{5}

Hence, the slope of the line should be 1/5.

The line passes through (-2,2).

Now use point slope form to find the equation of line.

y-y_1=m(x-x_1)

Substitute m=1/5, x₁=-2 and y₁ = 2 in above formula.

y-2=\frac{1}{5}(x+2)

y-2=\frac{x}{5}+\frac{2}{5}

y=\frac{x}{5}+\frac{2}{5}+2

y=\frac{x}{5}+\frac{12}{5}

Hence, the equation in slope-intercept form for the path of the trumpet players is y=\frac{x}{5}+\frac{12}{5}.

4 0
1 year ago
a group of girls bought 72 rainbow hairbands ,144 brown and black hairbands and 216 bright colored hairbands .what is the larges
Nitella [24]

Answer:

72

Step-by-step explanation:

We are assuming that all girls in the group bought the same number of items.

Therefore, we need to find the highest common factor of 72, 144 and 216.

HCF

72 = 2 * 2 * 2 * 3 * 3

144 = 2 * 2 * 2 * 2 * 3 * 3

216 = 2 * 2 * 2 * 3 * 3 * 3

The product of the emboldened numbers is the highest common factor.

That is:

2 * 2 * 2 * 3 * 3 = 72

Therefore, the largest possible number of girls in the group is 72.

5 0
2 years ago
A and B will take the same 10-question examination. Each question will be answered correctly by A with probability .7, independe
MatroZZZ [7]

Answer:

a) The expected number of questions that are answered correctly by both A and B = 11 (7 + 4).

b) The Variance of the number of questions that are answered correctly by either A or B = 2.25.

Step-by-step explanation:

Number of questions in the examination = 10

Probability of A's answer being correct = 0.7

Probability of B's answer being correct = 0.4

The expected number of questions that are answered correctly by both A and B:

           Probability of       Expected        

        Correct Answer       Value          Variance

A              0.7                     7 (0.7 * 10)    2.25

B              0.4                     4 (0.4 * 10)    2.25

Total expected value =    11

           Mean =                  5.5                2.25

5 0
1 year ago
Two professors are applying for grants. Professor Jane has a probability of 0.61 of being funded. Professor Joe has probability
Anna11 [10]

Answer:

  • a. 0.1647;
  • b. 0.7153;
  • c. 0.4453.

Step-by-step explanation:

By definition, if two event A and B are independent, then

P(A\cap B) = P(A) \cdot P(B). (P(A\cap B) is the probability that the outcome of both event A and event B are true.)

<h3>a.</h3>

Since the outcome of these two events are independent,

\begin{aligned}P(\texttt{Jane} \cap \texttt{Joe}) &= P(\texttt{Jane}) \cdot P(\texttt{Joe})\\ &= 0.61 \times 0.27 \\&= 0.1647\end{aligned}.

<h3>b.</h3>

The logic not operator \lnot or the prime superscript ^{\prime} denotes that an event does not happen.  

P(\texttt{Jane}^{\prime}) = 1 - P(\texttt{Jane}) = 1- 0.61 = 0.39.

P(\texttt{Joe}^{\prime}) = 1 - P(\texttt{Joe}) = 1- 0.27 = 0.73.

Since the two events \texttt{Jane} and \texttt{Joe}, \texttt{Jane}^{\prime} and \texttt{Joe}^{\prime} are also independent. Probability that neither professor got funded:

P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = P(\texttt{Jane}^{\prime}) \cdot P(\texttt{Joe}^{\prime}) = 0.39 \times 0.73 = 0.2847.

Probability that at least one professor got funded- in other words, it is not true that neither professor got funded:

P((\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime})^{\prime}) = 1- P(\texttt{Jane}^{\prime} \cap \texttt{Joe}^{\prime}) = 0.7153.

<h3>c.</h3>

Similarly, since the two events \texttt{Jane} and \texttt{Joe}, \texttt{Jane} and \texttt{Joe}^{\prime} are also independent. Probability that Jane but not Joe got funded:

P(\texttt{Jane} \cap (\texttt{Joe}^{\prime})) = P(\texttt{Jane}) \cdot P(\texttt{Joe}^{\prime}) = 0.61 \times 0.73 = 0.4453.

4 0
1 year ago
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