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vampirchik [111]
2 years ago
9

Easy question and it's a warm up question. Noah has to do 8 homework and he did 7 homework. How many homework left does noah hav

e to do? Explain how you got the answer.
Mathematics
2 answers:
nadezda [96]2 years ago
4 0

Answer:

1

Step-by-step explanation:

(H - Homework)

H = 8

H = 8 - 7

H = 1

UNO [17]2 years ago
4 0

Answer:

1 assignment left

Step-by-step explanation:

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Factor. (help plz!)<br><br> x^2+7x+10
yawa3891 [41]

Answer:

(x + 5) \times (x + 2)

Step-by-step explanation:

{x}^{2}  + 7x + 10

{x}^{2}  + 5x + 2x + 10

x \times (x + 5) + 2(x + 5)

(x + 5) \times (x + 2)

3 0
3 years ago
Read 2 more answers
Two jets leave an airport at the same time, flying in opposite directions. The first jet is traveling at three hundred seventy-s
gladu [14]
Here is the set using D = rt.


Jet 1:

rate = 377 mph

time = x

Distance = 9128 miles

Jet 2:

rate = 275 mph

time = x

Distance = 9128 miles

Equation:

377x + 275x = 9128

You finish.
8 0
3 years ago
We have five samples of data: sample A with 30 successes of 50 cases, sample B with 600 successes of 1000 cases, sample C with 3
faust18 [17]

Answer:

C. with 3000 successes of 5000 cases sample

Step-by-step explanation:

Given that we need to test if the proportion of success is greater than 0.5.

From the given options, we can see that they all have the same proportion which equals to;

Proportion p = 30/50 = 600/1000 = 0.6

p = 0.6

But we can notice that the number of samples in each case is different.

Test statistic z score can be calculated with the formula below;

z = (p^−po)/√{po(1−po)/n}

Where,

z= Test statistics

n = Sample size

po = Null hypothesized value

p^ = Observed proportion

Since all other variables are the same for all the cases except sample size, from the formula for the test statistics we can see that the higher the value of sample size (n) the higher the test statistics (z) and the highest z gives the strongest evidence for the alternative hypothesis. So the option with the highest sample size gives the strongest evidence for the alternative hypothesis.

Therefore, option C with sample size 5000 and proportion 0.6 has the highest sample size. Hence, option C gives the strongest evidence for the alternative hypothesis

3 0
3 years ago
A group of 54 college students from a certain liberal arts college were randomly sampled and asked about the number of alcoholic
Ket [755]

Answer:

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

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

p_v =2*P(t_{53}>0.9601)=0.3414  

Step-by-step explanation:

Data given and notation  

\bar X=5.25 represent the sample mean

s=3.98 represent the sample standard deviation

n=54 sample size  

\mu_o =4.73 represent the value that we want to test  

\alpha represent the significance level for the hypothesis test.  

z 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 mean differs from 4.73, the system of hypothesis would be:  

Null hypothesis:\mu =4.73  

Alternative hypothesis:\mu \neq 4.73  

Since we know don't the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic  

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

t=\frac{5.25-4.73}{\frac{3.98}{\sqrt{54}}}=0.9601  

P-value  

First we find the degrees of freedom given by:

df = n-1= 54-1=53

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

p_v =2*P(t_{53}>0.9601)=0.3414  

7 0
3 years ago
The point (-7,6) is in the
san4es73 [151]

Answer:

need a pic to see

Step-by-step explanation:

8 0
3 years ago
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