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Aneli [31]
3 years ago
8

PLEASE ANSWER ASAP For the Christmas and holiday season, every student in Mrs. Perri’s class decided to send a card to each stud

ent in the class and to Mrs. Perri. There are 31 students in the class. How many cards were sent that day?
Mathematics
1 answer:
MaRussiya [10]3 years ago
4 0
Each student sent 32 cards.

There are 31 students in the class.

To find the total number of cards sent, multiply 32 by 31.
32x31 = 992

992 cards were sent that day.
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The best possible first step would be to multiply equation one by 4 and equation by 5 that would eliminate the y and allow you to solve for x.

In essence,      4 × (3x  +  5y = 1)   eq1
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   thus giving rise to  12x + 20y = 4     eq1 × 4
                                  10x +20y = -20  eq2 × 5

You can then just subtract one equation from the other
For example,   (12x + 20y = 4) - ( 10x +20y = -20 ) 

                        12x - 10x + 20y - 20y = 4 - (-20) 
                                                      2x = 24
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8 0
3 years ago
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A company advertises two car tire models. The number of thousands of miles that the standard model tires last has a mean μ S​ =6
Lera25 [3.4K]

Answer: .123

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Christine went shopping and bought each of her six nephews a gift, either a video costing $14.95 for a CD costing $16.88. She sp
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Let v = number of video's she bought

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DedPeter [7]

Answer:

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Hope this helps!

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3 years ago
Pilates is a popular set of exercises for the treatment of individuals with lower back pain. The method has six basic principles
mariarad [96]

Answer:

p_v =P(z>2.087) =0.0184

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.  

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

\sigma=2.5

And the statistic is given by this formula:

z=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sigma\sqrt{\frac{1}{n_1}}+\frac{1}{n_2}}

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Null hypothesis: \mu_c \leq \mu_t+1

Alternative hypothesis: \mu_c >\mu_t+1

Or equivalently:

Null hypothesis: \mu_c - \mu_t \leq 1

Alternative hypothesis: \mu_c-\mu_t>1

Our notation on this case :

n_c =45 represent the sample size for group control

n_t =45 represent the sample size for group  treatment

\bar X_c =5.2 represent the sample mean for the group control

\bar X_t =3.1 represent the sample mean for the group treatment

And now we can calculate the statistic:

z=\frac{(5.2-3.1)-(1)}{2.5\sqrt{\frac{1}{45}}+\frac{1}{45}}=2.087  

And now we can calculate the p value using the alternative hypothesis:

p_v =P(z>2.087) =0.0184

So with the p value obtained and using the significance level given \alpha=0.01 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 1% of significance the mean for the control group it's not significantly higher than the mean of the treatment group by 1 point.  

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