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Gnesinka [82]
3 years ago
9

I need help with this?

Mathematics
1 answer:
Anastaziya [24]3 years ago
6 0

Answer:

it is 7

Step-by-step explanation:

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Please Answer and Show your Work: <br><br> 4x/16 - 3/8 - x/16 = 3/16
Galina-37 [17]
4x/16 - 3/8 - x/16 = 3/16

x/4 - 3/8 - x/16 = 3/16

3x/16 - 3/8 = 3/16

Add 3/8 on both sides

3x/16 = 9/16

Then multiply 16 on both sides

3x = 144/16

3x = 9

Divide 3 on both sides

x = 3
8 0
3 years ago
Outcomes are equally likely of each is as likely to occur true or false
Brums [2.3K]

Answer:

true

Step-by-step explanation:

8 0
3 years ago
If anyone could answer this for me I would very much appreciate it
baherus [9]

Answer:

a3 = 45

Step-by-step explanation:

a1 = 20

an = a(n-1) *3/2  where n is the term number

a2 = a(2-1) *3/2

     a1 *3/2

We know a1 = 20

a2 = 20*3/2 = 30

a3 = a(3-1) *3/2

    = a2 *3/2

We know a2 = 30

    = 30*3/2 = 45

8 0
4 years ago
Using the formula below, calculate the z-score for the listed data points. z-score: zx = x − μ σz1= z5=z6.5=
Pani-rosa [81]

<span>z1 = </span> -1.95

<span>z5 = </span> 0.05

<span>z6.5 = 0.8</span>

7 0
3 years ago
Read 2 more answers
A horticulturist working for a large plant nursery is conducting experiments on the growth rate of a new shrub. Based on previou
sergejj [24]

Answer:

t=\frac{2.9-3}{\frac{0.3}{\sqrt{50}}}=-2.357    

df=n-1=50-1=49  

p_v =P(t_{(49)}  

Since the p value is less than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true growth rate is significantly less than 3 cm per week

Step-by-step explanation:

Information given

\bar X=2.90 represent the sample mean for the growth

s=0.30 represent the sample standard deviation

n=50 sample size  

\mu_o =3 represent the value to check

\alpha=0.1 represent the significance level

t would represent the statistic

p_v represent the p value for the test

System of hypothesis

We want to verify if the true mean for the growth us less than 3cm per week, the system of hypothesis are:  

Null hypothesis:\mu \geq 3  

Alternative hypothesis:\mu < 3  

We don't know the population deviation so we can use the t statistic:

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

Replacing the info given we got:

t=\frac{2.9-3}{\frac{0.3}{\sqrt{50}}}=-2.357    

The degrees of freedom are given by:

df=n-1=50-1=49  

The p value for this case would be given by:

p_v =P(t_{(49)}  

Since the p value is less than the significance level of 0.1 we have enough evidence to reject the null hypothesis and we can conclude that the true growth rate is significantly less than 3 cm per week

3 0
3 years ago
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