Solve each system of equations without graphing.

Zero product property: (x-4)(-5x+1)=0 lesser x= greater x=

GaryK [48]

Answer:

lesser x= (x-4)greater x=(-5x+1)

5 0

3 years ago

What is the length of JM in the given figure?

umka21 [38]

Answer: B. 30

Step-by-step explanation:

When given a secant and a tangent, the formula is:

exterior of secant × secant = tangent²

KM × JK = LK²

10 × (JM + 10) = 20²

10JM + 100 = 400

10JM = 300

JM = 30

7 0

3 years ago

Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8

Delvig [45]

We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51$

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22$

Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98$

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = $m-z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Lower end of confidence interval= $123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34$
Upper end of confidence interval = $m+z *\frac{psd}{ \sqrt{n} }$
Using the values, we get:
Upper end of confidence interval= $123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86$

Thus the 98% confidence interval will be (105.34, 141.86)

Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.

3 0

4 years ago

The formula for the area of a rhombus is A = a equals StartFraction one-half EndFraction d 1 d 2.D1d2, where d1 and d2 are the l

Gelneren [198K]

Answer:

d1=2A/d2

d2=2A/d1

Step-by-step explanation:

The options of this question are:

1. d₁=2Ad₂

2. d₁= 2A/d₂

3. d₂= d₁/2A

4. d₁= 2A/d₂

5. d₂= 2Ad₁

Given:

A=1/2(d1*d2)

Multiply both sides by 2

We have,

2A=d1*d2

Divide both sides by d2

2A/d2=d1*d2/d2

2A/d2=d1

Therefore, d1=2A/d2

Similarly, from the previous equation

2A=d1*d2

Divide both sides by d1

2A/d1=d1*d2/d1

2A/d1=d2

Therefore,

d2=2A/d1

Options

2. d₁= 2A/d₂

4. d₁= 2A/d₂

7 0

3 years ago

Help!!!!!!!!!.!.!.!.!.

m_a_m_a [10]

Answer:

24

Step-by-step explanation:

4 0

3 years ago