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Sliva [168]
4 years ago
13

9. A car has an average speed of 80 km/h for one hour, then an

Mathematics
1 answer:
dsp734 years ago
7 0

Answer:

86.67 km/hr

Step-by-step explanation:

First Hour:  (80 km/hr)(1 hr)             =  80 km

Next two hours:  (90 km/hr)(2 hr)   =  180 km

Total distance traveled:                   = 260 km

Average speed:  total distance / total time elapsed) = 260 km  /  3 hr

= 86.67 km/hr

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Find a linear equation whose graph is the straight line with the given property.
jeka57 [31]

Answer:

y = 3/4x - 9/8

Then read explanation and see here with 2 coordinates would make

if 3/4 = -6 then 1 = -8

y = -6x + c

-6 / 3/4  + 9/8 / - 0.5  = -2.25

therefore c = -2.25

y = -6x - 2.25

y = -6x - 9/4  and again 9/4 can be found easily if round up fraction decimal to 225/100 and divide by 4

Step-by-step explanation:

C1)    (0.5 , -0.75)

C2)   ( 1. -3.75)

y = mx + c

-3.75 - - 0.75  /   1 - 0.5  = -3.75 + 0.75 / 0.5

= -3 / 0.5 = m

m = -6

y = -6x + c

then this works given just one coordinate to solve for 1 or 2 sets given points

y - y1 = mx+c

y - - 0.75 =  0.75 (x - x1)

y + 0.75 = 0.75 (x - 0.5)

y + 0.75 = 0.75x - 0.375

y = 0.75 - 0.75   =  0.75x( - 0.375 - 0.75)

y = 0.75x -1.125

y = 3/4x - 9/8    as common denominators of 1125/1000 is 9/8 as we divide by  125 into the fraction as we start at 250 and check its half just like when finding common denominators we 1/4 the lower number and see if we can find its half or its double etc with the other num er as first step for 3sf numbers.

5 0
3 years ago
Muffins74blueberry muffins7banana nut muffins6pumpkin spice muffins11
BartSMP [9]
6/74 i think. So #banananutmuffin/#muffinsintotal
6 0
3 years ago
Two students were asked if they liked to read. Is this a good example of a statistical question? Why or why not? A Yes, this is
guapka [62]

Answer:

c

Step-by-step explanation:

only 2 students were asked the question, 2 students don't represent the whole population

3 0
3 years ago
Read 2 more answers
The article "Students Increasingly Turn to Credit Cards" (San Luis Obispo Tribune, July 21, 2006) reported that 37% of college f
Sloan [31]

Answer:

Step-by-step explanation:

Hello!

There are two variables of interest:

X₁: number of college freshmen that carry a credit card balance.

n₁= 1000

p'₁= 0.37

X₂: number of college seniors that carry a credit card balance.

n₂= 1000

p'₂= 0.48

a. You need to construct a 90% CI for the proportion of freshmen  who carry a credit card balance.

The formula for the interval is:

p'₁±Z_{1-\alpha /2}*\sqrt{\frac{p'_1(1-p'_1)}{n_1} }

Z_{1-\alpha /2}= Z_{0.95}= 1.648

0.37±1.648*\sqrt{\frac{0.37*0.63}{1000} }

0.37±1.648*0.015

[0.35;0.39]

With a confidence level of 90%, you'd expect that the interval [0.35;0.39] contains the proportion of college freshmen students that carry a credit card balance.

b. In this item, you have to estimate the proportion of senior students that carry a credit card balance. Since we work with the standard normal approximation and the same confidence level, the Z value is the same: 1.648

The formula for this interval is

p'₂±Z_{1-\alpha /2}*\sqrt{\frac{p'_2(1-p'_2)}{n_2} }

0.48±1.648* \sqrt{\frac{0.48*0.52}{1000} }

0.48±1.648*0.016

[0.45;0.51]

With a confidence level of 90%, you'd expect that the interval [0.45;0.51] contains the proportion of college seniors that carry a credit card balance.

c. The difference between the width two 90% confidence intervals is given by the standard deviation of each sample.

Freshmen: \sqrt{\frac{p'_1(1-p'_1)}{n_1} } = \sqrt{\frac{0.37*0.63}{1000} } = 0.01527 = 0.015

Seniors: \sqrt{\frac{p'_2(1-p'_2)}{n_2} } = \sqrt{\frac{0.48*0.52}{1000} }= 0.01579 = 0.016

The interval corresponding to the senior students has a greater standard deviation than the interval corresponding to the freshmen students, that is why the amplitude of its interval is greater.

8 0
3 years ago
Cos inverse(3/5 cos x+4/5 sin x)
Vesnalui [34]
3cos (x) / 5 + 4sin (x)/ 5 

this is simplified
6 0
4 years ago
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