HELP PLEASE!!! Answer: ?

What is the measure of ABC? A. 130 B. 85 C.40 D.170

Nitella [24]

Answer:

130

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the answer for “round 756,04 to the nearest tenth

777dan777 [17]

If you mean 756.04, the answer is 756.0. 4 is closer to 0.

5 Or More, raise the score
4 Or Less, let it rest

7 0

3 years ago

a plane is flying at a speed of 200 miles per hour. the pilot plans on flying at this speed for the next 225 miles, plus or minu

harina [27]

The minimum and maximum number of hours the plane will travel at that speed are; at least 0.875 hours and at most 1.375 hours

<h3>How to Solve Absolute Value Inequality?</h3>

We are given that the speed of the plane is 200 miles per hour.

Now, formula for time taken is;

Time = Distance/Speed

If the speed is 225 ± 50 miles, then it means that;

Maximum distance = 275 miles

Minimum distance = 175 miles

Thus, we can say that;

Minimum number of hours = 275/200 = 1.375 hours

Maximum number of hours = 175/200 = 0.875 hours

Read more about Absolute Value Inequality at; brainly.com/question/13282457

#SPJ1

4 0

2 years ago

Jimmy’s new cell phone cost him $49.99 when he signed a 2 year plan, which was 75% off the original place. What was the original

Sati [7]

If Jimmy's new cell cost him $49.99 with a 75% discount, then $49.99 is 100%−75%=25% of the original price.

The original price of the cell phone was $199.96.

6 0

4 years ago

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores

EastWind [94]

Answer:

$(52-49) -4.300= -1.300$

$(52-49) +4.300= 7.300$

And the 95% confidence would be :

$-1.300 \leq \mu_1 -\mu_2 \leq 7.300$

Step-by-step explanation:

We have the following info given from the problem

$\bar X_1 = 52$ sample mean for this year

$s_1= 13$ sample deviation for this year

$n_1 = 75$ random sample selected for this year

$\bar X_2 = 49$ sample mean for last year

$s_2= 11$ sample deviation for last year

$n_1 = 53$ random sample selected for last year

And we want to construct a 95% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales

For this case the formula that we need to use is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$