The two equations are similar because they both end up with the same common value. 0=0 The work you need to show that this is correct will be shown below.

Step-by-step explanation:

2(5m−4)−3(m−5)2=−3m2+40m−83

Add 3m^2 to both sides.

−3m2+40m−83+3m2=−3m2+40m−83+3m2

40m−83=40m−83

Subtract 40m from both sides.

40m−83−40m=40m−83−40m

−83=−83

−83+83=−83+83

0=0

If you have any questions regarding my answer please tell me in the comments, I will come and answer them. Have a good day.

6 0

2 years ago

Really need help :( struggling

Scorpion4ik [409]

Answer:

Step-by-step explanation:

The original amount will be the coefficient of (1 + 0.04)^r which is $728.00.

6 0

2 years ago

A buyer went to the market to buy strawberries. He purchased 120 randomly selected strawberries from a vendor who claimed that n

mafiozo [28]

Answer:

$z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10$

$p_v =P(z>2.10)=0.018$

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

Step-by-step explanation:

Data given and notation

n=120 represent the random sample taken

X=40 represent the number of strawberries damaged

$\hat p=\frac{40}{120}=0.333$ estimated proportion of strawberries damaged

$p_o=0.25$ is the value that we want to test

$\alpha$ represent the significance level

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that no more than 25% of his total harvest of strawberries was damaged.:

Null hypothesis: $p\leq 0.25$

Alternative hypothesis: $p > 0.25$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.333 -0.25}{\sqrt{\frac{0.25(1-0.25)}{120}}}=2.10$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>2.10)=0.018$

At 5% of significance we can conclude that the true proportion of strawberries damage is higher than 0.25

6 0

2 years ago

It is now 8:59 a.m. when the bell rings at 9:01 am. Damian will be late for science class for the 3rd time this week. He must ge

kvv77 [185]

Answer:

1: 10 seconds 2: 4 seconds 3: 12 seconds

Step-by-step explanation:

35.0/3.5= 10

4.80/1.20= 4

60.0/5.00= 12

4 0

2 years ago