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ICE Princess25 [194]
3 years ago
15

PLZ HELP!!!

Mathematics
2 answers:
bagirrra123 [75]3 years ago
5 0
B) because the rate of change is 7, while for A the rate of change is 6 and for C and D the rate of change is 4
Xelga [282]3 years ago
4 0
Answer is B because is make the add is answer
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Reflect shape A in the line y = x.
marshall27 [118]

Answer:

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8 0
2 years ago
Nicole is playing a video game where each round lasts \dfrac{7}{12} 12 7 ​ start fraction, 7, divided by, 12, end fraction of an
Harrizon [31]

Total number of hours she schduled to play the game = 3\frac{3}{4} hours.

Let us convert mixed fraction into improper fraction

3\frac{3}{4} = \frac{3*3+4}{4} = \frac{13}{4} \ hours.

Duration of each round = \frac{7}{12}.

In order to find the number of rounds Nicole can play, we need to divide total number of hours by duration of each round.

\frac{13}{4} ÷ \frac{7}{12}

Converting division sign into multiplication flips the second fraction.

=\frac{13}{4} \times \frac{12}{7}

Crossing out 12 by 4, we get 3 on the top of second fraction.

=\frac{13}{1} \times \frac{3}{7}

=\frac{39}{7}  = 5.57...( Approximately).

Because problem is about number of rounds.

So, total number of round would be 5.

5 0
3 years ago
Read 2 more answers
Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From
maksim [4K]

Answer:

The 95% CI for the difference of means is:

-155.45 \leq \mu_1-\mu_2 \leq -44.55

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size:  4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

M_d=M_1-M_2=200-300=-100

The standard deviation can be estimated as:

\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45

The degrees of freedom are:

df=n_1+n_2-2=10+4-2=12

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55

8 0
3 years ago
A negative number raised to an odd power is _____ negative.
BartSMP [9]
Answer – Always
 A negative number raised to an odd power is always negative. When a negative number is raised to an even power, the pairs of negatives will cancel out; but when it is raised to an odd power, after pairs of the negative sign have canceled each other out, there will still be one minus sign left unpaired, which will not cancel out. <span>
</span>
4 0
3 years ago
Read 2 more answers
Jaz has the following scores from a new game:
Marina86 [1]

Answer:

the mean of it all will increase

Step-by-step explanation:

3 0
2 years ago
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