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stepladder [879]
3 years ago
10

Round the decimal to the nearest hundreth. If decimal cannot be rounded, state this. 194.59

Mathematics
2 answers:
sveta [45]3 years ago
4 0
The answer is 194.60
notsponge [240]3 years ago
4 0

The decimal cannot be rounded. It is already rounded to the nearest hundredth.

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Pls answer as soon as possible what is 5.3 rounded?
Nadusha1986 [10]

Answer:

5

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Jenna saves $3 for every $13 dollars she earns. Vanessa saves $6 for every $16 she earns. Is Jenna's ratio of money saved to mon
jeka94

We have been given that Jenna saves $3 for every $13 dollars she earns. Vanessa saves $6 for every $16 she earns.

We can compare the ratios of saved money to earned money by Jenna and Vanessa as:

\frac{3}{13} =\frac{6}{16}    

Let us simplify our equation.

0.2307692307692308=0.375    

We can see that 0.231\neq 0.375 , therefore, Jenna's ratio of money saved to money earned is not equivalent to Vanessa's ratio of money saved to money earned.


5 0
3 years ago
A vet treats dogs (D), cats (C), birds (B), hamsters (H), and reptiles (R). A veterinary assistant randomly selects a patient’s
Kryger [21]
S={D, C, B, H, R}

Hope this helps

7 0
4 years ago
Read 2 more answers
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).  

7 0
4 years ago
Solve for h.<br><br> –2 − 10h = –11h − 20
Delvig [45]

。☆✼★ ━━━━━━━━━━━━━━  ☾  

-2 - 10h = -11h - 20

+ 2

- 10h = -11h - 18

+ 11h

h = -18

Have A Nice Day ❤    

Stay Brainly! ヅ    

- Ally ✧    

。☆✼★ ━━━━━━━━━━━━━━  ☾

5 0
3 years ago
Read 2 more answers
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