Help me with the answer

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Alex

log₇ (50) - log₇ (5) = log₇ (50 ÷ 5)

log₇ ( $\frac{50}{5}$ ) = log₇ (10)

Answer: C

8 0

4 years ago

Using the Distributive Property to factorize the equation 3x2 + 24x = 0, you get . The solution of the equation is

coldgirl [10]

3x(x+8)=0
x=0,-8
This is how to solve for x. Comment if you need something else.

7 0

3 years ago

Read 2 more answers

If you invested $600 at 3% simple interest for 2 years, how much interest do you earn? Show work and answer in complete sentence

zzz [600]

Answer:

I would rather do the second option of which uses Compound interest that will give a profit of $47.85

Step-by-step explanation:

In this problem we will be exploring the two formulas

1. simple interest

A= P(1+r*t)

2. compound interest

A= P(1+r/n)^nt

Where A= final amount

P= initial amount

r= rate

t= time.

n= number of periods Compounded

1.given data

P= $600

r= 3%= 3/100= 0.03

t= 2 years

A= 600(1+0.03*2)

A= 600(1+0.06)

A= 600(1.06)

A= $636

Interest = 636-600= $36

2. Given data

P= $600

r= 4%= 4/100= 0.04

n= 24

t= 2

A= 600(1+0.04/24)^24*2

A=600(1+0.0016)^48

A=600(1.0016)^48

A= 600*1.07975

A= 647.85

Interest = 647.85-600= $47.85

7 0

3 years ago

The germination rate is the rate at which plants begin to grow after the seed is planted.

chubhunter [2.5K]

Answer:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

Step-by-step explanation:

Data given and notation

n=15 represent the random sample taken

X=7 represent the number of seeds germinated

$\hat p=\frac{7}{15}=0.467$ estimated proportion of seeds germinated

$p_o=0.9$ 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 the true proportion of germinated seeds is less than 0.9 or 90%.:

Null hypothesis: $p\geq 0.9$

Alternative hypothesis: $p < 0.9$

When we conduct a proportion test we need to use the z statisitc, 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:

$z=\frac{0.467 -0.9}{\sqrt{\frac{0.9(1-0.9)}{15}}}=-5.59$

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 significance level assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of germinated seeds is significantly lower than 0.9 or 90%

4 0

3 years ago

Three lightship flash simultaneously at 6 00 a.m the first lightship flashes every 12 sec the second lightship every 30 sec and

Dmitrij [34]

They will all flash their lights together at 5:00 pm

6 0

3 years ago