The ratio of the number of people at a dinner to the number invited was 106 to 100. Express this as a percent

Answer:
1) correct
2) correct
3) Not Congruent
4) Not Congruent
5) correct
6) Not Congruent
7) incorrect
8) SAS

Explanation:
7 is incorrect because the angle isn’t included in the sides

Hope this helps!

7 0

3 years ago

Karl drove 619.5 miles. For each gallon of gas, the car can travel 41 miles. What is a reasonable estimate of the number of gall

dusya [7]

Answer:

15

Step-by-step explanation:

Calculation for What is a reasonable estimate of the number of gallons of gas Karl used

Estimated number of gallons used=619.5 miles/41 miles

Estimated number of gallons used=15

Therefore the reasonable estimate of the number of gallons of gas Karl used is 15

6 0

3 years ago

The mean SAT score in mathematics, M, is 600. The standard deviation of these scores is 48. A special preparation course claims

kupik [55]

Answer:

Step-by-step explanation:

The mean SAT score is $\mu=600$ , we are going to call it \mu since it's the "true" mean

The standard deviation (we are going to call it $\sigma$ ) is

$\sigma=48$

Next they draw a random sample of n=70 students, and they got a mean score (denoted by $\bar x$ ) of $\bar x=613$

The test then boils down to the question if the score of 613 obtained by the students in the sample is statistically bigger that the "true" mean of 600.

- So the Null Hypothesis $H_0:\bar x \geq \mu$

- The alternative would be then the opposite $H_0:\bar x < \mu$

The test statistic for this type of test takes the form

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}$

and this test statistic follows a normal distribution. This last part is quite important because it will tell us where to look for the critical value. The problem ask for a 0.05 significance level. Looking at the normal distribution table, the critical value that leaves .05% in the upper tail is 1.645.

With this we can then replace the values in the test statistic and compare it to the critical value of 1.645.

$t=\frac{| \mu -\bar x |} {\sigma/\sqrt{n}}\\\\= \frac{| 600-613 |}{48/\sqrt(70}}\\\\= \frac{| 13 |}{48/8.367}\\\\= \frac{| 13 |}{5.737}\\\\=2.266\\$

<h3>since 2.266>1.645 we can reject the null hypothesis.</h3>

6 0

3 years ago

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