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2 years ago

PLEASE HELP I HAVE 3 MINS LEFT

Mathematics

1 answer:

lyudmila [28]2 years ago

5 0

I believe it is 5 or 10

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Solve the quadratic equations

Alexxx [7]

Answer:

for quadratic equation 1

y^2-10y+21=0

Step-by-step explanation:

y^2-7y-3y+21=0

y(y-7)-3(y-7)=0

(y-7)(y-3)=0

y-7=0,y-3=0

y=7,y=3

y=(3,7)

for quadratic equation 2

16p^2-8p+1=0

16p^2-4p-4p+1=0

4p(4p-1)-1(4p-1)=0

(4p-1)twice=0

4p-1=0,4p-1=0

4p=1

p=1/4 twice

for quadratic equation 3

x^2-400=0

x^2=400

x=√400

x=20

for quadratic equation 4

-16m^2-8m-1=0

multiply the equation by -

16m^2+8m+1=0

16m^2+4m+4m+1=0

4m(4m+1)1(4m+1)=0

4m+1=0 twice

m=-1/4 twice

for quadratic equation 5

-3n^2+75=0

divide both side by -3

-3n^2/-3=-75/-3

n^2=25

n=√25

n=5

6 0

2 years ago

A study of 35 golfers showed that their average score on a particular course was 92. The sample standard deviation was 5. We wan

svlad2 [7]

Answer:

$t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$

Step-by-step explanation:

We have a sample of size n = 35, $\bar{x} = 92$ and $s = 5$ . As we want to construct a 95% confidence interval to estimate the value of the true population mean $\mu$ , we should use the pivotal quantity given by $T = \frac{\bar{X}-\mu}{S/\sqrt{n}}$ which comes from the t distribution with n - 1 = 35 - 1 = 34 degrees of freedom. Besides we should use the t-value $t_{\alpha/2} = t_{0.05/2} = t_{0.025} = 2.0322$ , i.e., regarding the t-distribution with 34 degrees of freedom, we have an area of 0.025 above 2.0322 and below the probability density function. The rationale behind this is that $P(-2.0322\leq T\leq 2.0322) = 0.95$ because of the simmetry of the t distribution.