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Anna11 [10]
3 years ago
6

ANSERR GIVE ME ANSWERS PLS

Mathematics
1 answer:
Zarrin [17]3 years ago
4 0

Answer:

Concept: Mathematics

When you multiply a number by something to the 10 power of something then you move whatever place of the exponential to the right if its positive and left if it's negative.

  1. 1.365 × 10^1
  2. You see its positive 1 hence we move one space to the right
  3. Our answer is now 13.65

You keep doing the same thing

  1. 1.365 × 10^2
  2. 136.5

Lastly

  1. 1.365 ×10^3
  2. 1365

Rate positively and give 5 stars

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A medical researcher wondered if there is a significant difference between the mean birth weight of boy and girl babies. Random
Yuliya22 [10]

Answer:

b) independent samples t-test

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 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 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

Step-by-step explanation:

The system of hypothesis on this case are:  

Null hypothesis: \mu_1 = \mu_2  

Alternative hypothesis: \mu_1 \neq \mu_2  

Or equivalently:  

Null hypothesis: \mu_1 - \mu_2 = 0  

Alternative hypothesis: \mu_1 -\mu_2\neq 0  

Our notation on this case :  

n_1 =5 represent the sample size for group 1  

n_2 =5 represent the sample size for group 2  

We can calculate the mean and deviation for eaach group with the following formulas:

\bar X= \frac{\sum_{i=1}^n X_i}{n}

s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

\bar X_1 =5.92 represent the sample mean for the group 1  

\bar X_2 =5.74 represent the sample mean for the group 2  

s_1=1.88 represent the sample standard deviation for group 1  

s_2=1.81 represent the sample standard deviation for group 2  

If we see the alternative hypothesis we see that we are conducting a bilateral test and with independnet samples t test.

b) independent samples t-test

The statistic is given by this formula:  

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{S^2_2}{n_2}}  

And now we can calculate the statistic:  

t=\frac{(5.92 -5.74)-(0)}{\sqrt{\frac{1.88^2}{5}}+\frac{1.81^2}{5}}=0.154  

The degrees of freedom are given by:  

df=5+5-2=8

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

p_v =2*P(t_{8}>0.154) =0.881  

So with the p value obtained and using the significance level given \alpha=0.05 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 5% of significance the mean of the group 1 (boys) is NOT significantly different than the mean for the group 2 (Girls).  

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Simplify by combining like terms: 9x - 3y -5x + 2y -x + 2
UNO [17]

Answer:

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Step-by-step explanation:

9x-5x-x-3y+2y=-2

3x-y=-2

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pochemuha

Answer:

nooooooo

Step-by-step explanation:

nooooo. vhjggbbb

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