Triangle MNO is similar to triangle RPO.

What additional information could be used to prove ΔABC ≅ ΔMQR using SAS? Select two options. m∠A = 64° and AB = MQ = 31 cm CB =

enyata [817]

Answer:

m∠R = 60° and AB ≅ MQ

m∠Q = 56° and CB ≅ RQ

Step-by-step explanation:

Given data :

Prove ΔABC ≅ ΔMQR using SAS

The missing information to prove ΔABC ≅ ΔMQR using SAS

m∠R = 60° and AB ≅ MQ

m∠Q = 56° and CB ≅ RQ

3 0

4 years ago

Read 2 more answers

A bookstore wants to determine how many books the people in the surrounding neighborhood read per month on average. they survey

GREYUIT [131]

Answer:

Step-by-step explanation:

poop

4 0

3 years ago

Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com

Ymorist [56]

Answer:

a) $183-1.984\frac{20}{\sqrt{100}}=179.032$

$183+1.984\frac{20}{\sqrt{100}}=186.968$

So on this case the 95% confidence interval would be given by (179.032;186.968)

b) $190-1.984\frac{23}{\sqrt{100}}=185.437$

$190+1.984\frac{23}{\sqrt{100}}=194.563$

So on this case the 95% confidence interval would be given by (185.437;194.563)

c) For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Part a : Summer

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=100-1=99$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that $t_{\alpha/2}=1.984$

Now we have everything in order to replace into formula (1):

$183-1.984\frac{20}{\sqrt{100}}=179.032$

$183+1.984\frac{20}{\sqrt{100}}=186.968$

So on this case the 95% confidence interval would be given by (179.032;186.968)

Part b: Winter

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=100-1=99$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,99)".And we see that $t_{\alpha/2}=1.984$

Now we have everything in order to replace into formula (1):

$190-1.984\frac{23}{\sqrt{100}}=185.437$

$190+1.984\frac{23}{\sqrt{100}}=194.563$

So on this case the 95% confidence interval would be given by (185.437;194.563)

Part c

For Summer the confidence interval was (179.032;186.968) and as we can see our upper limit is <190 so then we can conclude that they are below the specification of 190 at 5% of significance

For Winter the confidence interval was (185.437;194.563) and again the upper limit is <190 so then we can conclude that they are below the specification of 195 at 5% of significance

5 0

3 years ago

Electric wire: $1.49 for 3 ft. $.69 for 18 in.

vredina [299]

$1.49 3ft or $.69 1.5ft

69*2=138

$149 for 3ft or $1.49 1.5ft

options two are better buys

6 0

3 years ago

Can someone gift me the battle pass on fornite my user Emily_mari90

olga nikolaevna [1]

Answer:maybe

Step-by-step explanation:

took it

8 0

3 years ago