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Anna [14]
3 years ago
9

1100 students attend Ridgewood Junior High School. 4% of students bring their lunch to school everyday. How many students brough

t their lunch to school on Thursday?
Mathematics
2 answers:
Triss [41]3 years ago
7 0

Answer:

the answer is 44 students.

Step-by-step explanation:

valentinak56 [21]3 years ago
5 0

Answer: 44 students

Step-by-step explanation:

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LiRa [457]

Answer:

Hope this helps!

Step-by-step explanation:

2L + 3S = 83

 

4L + 8S = 197

 

Multiplies the top equation by -2.

-4L + -6S = -166

 

4l  + 8S = 197

 

 

Adding them together...

 

2S = 31

 

S = 31/2 = 15.5

 

Plugging into the first equation:

 

 2L + 3S = 83

 

 2L + 3(15.5) = 83

 

2L + 46.5 = 83

 

2L = 83 - 46.5

 

2L = 36.5

 

L = 36.5/2 = 18.25

 

check:

 

2L + 3S = 2(18.25) + 3(15.5) = 36.5 + 46.5 = 83

 

4L + 8S = 4(18.25) + 8(15.5) = 73 + 124 = 197

 

The Large box weighs 18.25

The small box weighs 15.5

6 0
2 years ago
Delia went to a market to buy 4.5 pounds of sliced fruit in a container she paid with a $20 bill and a $5 bill and received $4.6
andreyandreev [35.5K]
20+5=25 then 25-4.67=20.33 so then you do 20.33/4.5=4.5 so the cost for one pound of fruit is $4.57. Hope this helps :)
4 0
3 years ago
Let n = 5. What is the value of 10 + 15 ÷ n – 2? A. 3 B. 6 C. 11 D. 15 need help
Nataly_w [17]
11 hope dis helped u fam
7 0
3 years ago
Read 2 more answers
A forester studying diameter growth of red pine believes that the mean diameter growth will be different from the known mean gro
-Dominant- [34]

Answer:

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

Step-by-step explanation:

Data given and notation  

\bar X=1.6 represent the sample mean

s=0.46 represent the sample deviation

n=32 sample size  

\mu_o =1.35 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 1.35 in/year, the system of hypothesis would be:  

Null hypothesis:\mu =1.35  

Alternative hypothesis:\mu \neq 1.35  

The statistic is given by:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

t=\frac{1.6-1.35}{\frac{0.46}{\sqrt{32}}}=3.07  

P-value  

The degrees of freedom are given by:

df =n-1= 32-1=31

Since is a two-sided test the p value would be:  

p_v =2*P(t_{31}>3.07)=0.0044  

Since the p value is a very low value we have enough evidence to conclude that true mean is significantly different from 1.35 in/year at any significance level commonly used for example (\alpha=0.01,0.05, 0.1, 0.15).

5 0
3 years ago
Factor the quadratic expression Completely; -8x^2-15x+2
SSSSS [86.1K]
(x + 2)(- 8x + 1) are your factors. 
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3 years ago
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