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SpyIntel [72]
3 years ago
14

The scatterplot below shows the number of weeks that students have been in keyboarding class and the number of mistakes that the

y made on a keyboarding test. A graph titled Keyboarding mistakes has number of weeks in class on the x-axis and number of mistakes on the y-axis. Points are grouped together and decrease. Which statement about the scatterplot is true? As the number of weeks in class increases, the number of keyboarding mistakes decreases. As the number of weeks in class increases, the number of keyboarding mistakes increases. Regardless of the number of weeks in class, the number of keyboarding mistakes does not change. In general, decreased keyboarding mistakes do not affect the number of weeks in class.
Mathematics
2 answers:
Arlecino [84]3 years ago
6 0

Answer:A

Step-by-step explanation:

Took quiz

sergeinik [125]3 years ago
5 0

Answer:

A

Step-by-step explanation:

took the unit test

pls give brainliest

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Ship A and Ship B are 120 km apart when they pick up a distress call from another boat. Ship B estimates that they are 70 km awa
oksian1 [2.3K]

Answer:

147.5 km and 64.4 km

Step-by-step explanation:

a=120  km

b=70  km

β=28 degrees ( ∘)

 

b^2=(a^2)+(c^2)−2ac*cosβ  

70^2 =(120^2 )+(c^2)−2⋅ 120⋅ c⋅ cos(28∘ )  

 (c^2 ) −211.907c+9500=0  

 

note p, q, and r are replacement variables in the Pythagorean theorem since a, b, and c are already in use

p=1;q=−211.907;r=9500  

D=(q^2 ) −4pr=(211.907^2 )−4⋅1⋅9500=6904.75561996  

D>0  

 

c_{1,2}  =   (−q±  \sqrt{D}   )/2p=(211.91±\sqrt{6904.76})/2

​c_{1,2}  =105.95371114±41.5474295834  

(c_{1}−147.501140726)(c_{2}−64.4062815596)=0

c_{1}=147.501140726  

c_{2}=64.4062815596  

5 0
3 years ago
An electronics company has developed a new hand held device. The company predicts that the start up cost to manufacture the new
alex41 [277]
Profit = revenue - expenses

expenses : 125,000 + 6.50x
revenue : 9x
so to make a profit, ur revenue (income) has to be higher then ur expenses

revenue > expenses
A. ) 9x > 125,000 + 6.50x
9x - 6.50x > 125,000
2.5x > 125,000
x > 125,000 / 2.5
x > 50,000......so they would have to sell at least 50,001 devices to make a profit <==

B.) the cost of making 1 device is 10% more then the company predicted....10% more then 6.50.....6.50(1.10) = 7.15.....this is the new cost of making 1 device <==

9x > 125,000 + 7.15x ....this is the inequality with the 10% more added
9x - 7.15x > 125,000
1.85x > 125,000
x > 125,000 / 1.85
x > 67,567.5......so to make a profit, they would have to sell at least 67,568 devices to make a profit <==


7 0
3 years ago
What is the product?
sveticcg [70]

Answer:

third answer is correct

Step-by-step explanation:

hello,

we need to compute

5*(-1)+(-2)*(-1)=-5+2=-3

-6*(-1)+(2)*(-1)=6-2=4

so the correct answer is

   -3

    4

hope this helps

8 0
3 years ago
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
Please help it’s tomorrow at 10 am
Neporo4naja [7]

Step-by-step explanation:

Perimeter of swimming pool = 2(l + w)

2(24 + 12)

2(36)

72

Perimeter of walkway = 2(l + w)

2(4c + 4w) [ Since it is saying all side]

8c + 8w

Expression = 72 + 8c + 8w

8 0
3 years ago
Read 2 more answers
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