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3 years ago

Solve the system of linear equations.

Mathematics

1 answer:

RoseWind [281]3 years ago

4 0

Answer:

1. No solution

2. Infinite solutions

Step-by-step explanation:

1. To solve the system, set the equations equal to each other and solve for x.

2x - 5 = 2x + 7

-5 = 7

This is a false statement. This means there is no solution.

2. To solve the system, graph each equation.

y = -3/4 x - 5/2 has a y-intercept -5/2 and slope -3/4.

3x + 4y = -10 converts to y = -3/4x -5/2.

This graphs as the exact same line. This system has infinite solutions.

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Convert 76,000 ml into L. Please show your work.

tester [92]

1000 ml = 1 L

76000 ml
76*1000 ml
76 L

8 0

2 years ago

Read 2 more answers

Hi help meeeeeeee please i will mark you brainiest if you can answer this

Roman55 [17]

Answer:

Step-by-step explanation:

Honestly don't even care about the points lol. You need to divide 12^7 by 12^3, which comes out to 20,736 crates needed for the shipment of toys. If that first 12 isnt to the power of 3 simply replace it with the power that it actually is and divide 12^7 by it.

8 0

2 years ago

tia_tia [17]

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{s^2}{s^2}=\cfrac{18}{32}\implies \left( \cfrac{s}{s} \right)^2=\cfrac{18}{32}\implies \cfrac{s}{s}=\sqrt{\cfrac{18}{32}}
\\\\\\
\cfrac{s}{s}=\cfrac{\sqrt{18}}{\sqrt{32}}\implies \cfrac{s}{s}=\cfrac{\sqrt{9\cdot 2}}{\sqrt{16\cdot 2}}\implies \cfrac{s}{s}=\cfrac{\sqrt{3^2\cdot 2}}{\sqrt{4^2\cdot 2}}\implies \cfrac{s}{s}=\cfrac{3\sqrt{2}}{4\sqrt{2}}
\\\\\\
\cfrac{s}{s}=\cfrac{3}{2}$

6 0

3 years ago

An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 72 type K batteries and a sample of 49

rewona [7]

Answer:

Given:

For battery k(sample 1):

Sample size, n = 75

Sample mean, X' = 9.41

Standard deviation = 0.723

Mean u1 =???

For battery Q(sample 2):

Sample size, n = 49

Sample mean, X' = 9.69

Standard deviation = 0.257

Mean u2 =???

Significance level = 0.05

1) The null and alternative hypotheses, will be given as:

H0 : u1 - u2 = 0

H1 : u1 - u2 ≠ 0

2) This is a two tailed test.

For the test statistic, we have :

$Z_o = \frac{X'_1 - X'_2}{\sqrt{\frac{(\sigma_1)^2}{n_1} + \frac{(\sigma_2)^2}{n_2}}}$

$= \frac{9.41 - 9.69}{\sqrt{\frac{(0.723)^2}{75} + \frac{(0.257)^2}{49}}} = - 3.018$

≈ 3.02

The Zo = - 3.02

3) For the p value, using excel, we have:

2 * P(Z > (-3.018))

= 2 * NORMDIST (-3.018, 0, 1, true)

= 0.00254

For critical value:

Zcritical = Za/2 = 0.05/2

Zcritical = 0.025

We have alternative hypothesis,

H1 : u1 - u2 ≠ 0.

Therefore we reject null hypothesis H0 if Zo > Zcritical

4) We reject null hypothesis H0 if Zo > Zcritical. At 0.05 level of significance, there is not enough evidence to conclude that mean u1 and mean u2 are different.

4 0

2 years ago

Find the probability of drawing 3 Aces at random from a deck of 52 ordinary playing cards if the cards are:_________

Anuta_ua [19.1K]

Answer:

a. With replacement

1/2197

b. Without replacement

1/5,525

Step-by-step explanation:

Okay, here is a probability question.

The key to answering this question is by knowing the number of aces in a deck of cards.

There is 1 ace per suit, so there is a total of 4 aces per deck of cards.

So, mathematically the probability of picking an ace would be;

number of aces/ total number of cards = 4/52 = 1/13

a. Now since the action is with replacement; that means that at any point in time, the total number of cards would always remain 52 even after making our picks.

So the probability of picking three aces with replacement would be;

1/13 * 1/13 * 1/13 = 1/2197

b. Without replacement

what this action means is that after picking a particular card, we do not return the picked card to the deck of cards.

For the first card picked, we will be having a total of 4 aces and 52 total cards.

So the probability of picking an ace would be 4/52 = 1/13

For the second card picked, we shall be left with selecting an ace out of the remaining 3 aces and the total remaining 51 cards

So the probability will be 3/51 = 1/17

For the third and last card to be picked, we shall be left with picking 1 out of the remaining 2 aces cards and out of the 50 cards left in the deck.

So the probability now becomes 2/50 = 1/25

Thus, the combined probability of picking 3 aces cards without replacement from a deck of cards will be;

1/13 * 1/17 * 1/25 = 1/5,525

7 0

3 years ago