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Ivanshal [37]
3 years ago
7

Worth 20 points.

Mathematics
1 answer:
marshall27 [118]3 years ago
5 0
1.8(14)+25=50
after 14 months, paulette has 75 DVDs in her collection
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Graph y = x + 4<br><br> NEED HELP ASAP
lisabon 2012 [21]

Answer:

Step-by-step explanation:

x

y

−

4

0

−

3

1

−

3

−

1

−

2

2

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2

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4 0
3 years ago
PLEASE ANSWER ANY PROBLEM ASAP THANK YOU SO MUCH
Mashcka [7]

Answer:

1) No

2) Yes

3) angle UXS = 120 degrees

4) arc CB = 40 degrees

Step-by-step explanation:

If AB is tangent to the circle, then angle at A would be 90 degrees.

1)

Is 13^2 + 14.6^2 = 19.4^2?

169 + 213.16 = 376.36      No

2)

Is 8^2 + 6^2 = 10^2?

64 + 36 = 100      Yes

3)

arc WS = arc VU + arc TU = 113 degrees

arc ST = arc WV because their chords cross making a pair of vertical angles.

full circle = 360 degrees

360-113 - 113 = 134 degrees

134/2 = 67 degrees for arc ST

arc STU = 67 + 53 = 120 degrees

Angle UXS = 120 degrees

4) arc CB = arc FB =  40 degrees because their chords cross making a pair of vertical angles.

3 0
3 years ago
A student writes the equation for a line that has a slope of -6 and passes through the point (2, –8).
Tcecarenko [31]

Answer:

y = -6x + 20

Step-by-step explanation:

4 0
3 years ago
The standard recommendation for fluid needs is 30 ml/kg body weight. parker weighs 67 kg. how many cups of fluid will he need? (
Grace [21]
<span>Well First you multiply 67 kg, which is how much parker weighs, by 30. 30ml/kg means that there is 30ml for ever 1 kg and parker weighs 67. So Parker needs 2010 ml of fluid and the last thing you need to do is convert 2010 ml into cups. 1ml =.00423 cups. So you multiply 2010 by .00423 and you'd have you're answer. 8.5</span>
5 0
3 years ago
You would like to know whether forwards in a basketball league average the same (or different) numbers of points than the overal
Akimi4 [234]

Answer:

The critical t-value for such a test given an alpha level of 0.05 is 2.26

Step-by-step explanation:

Null hypothesis : H_0:\mu = 7.5

Alternate hypothesis :H_a:\mu \neq 7.5

Population mean = \mu = 7.5

Data : 10, 9, 6, 11, 13, 14, 9, 9, 9, and 10

Mean = \bar{x}=\frac{\text{Sum of all observations}}{\text{no. of observations}}

Mean =\frac{10+9+6+11+13+14+9+9+ 9+ 10}{10}

Mean =10

Standard deviation : \sqrt{\frac{\sum(x-\bar{x})^2}{n}}

Standard deviation :\sqrt{\frac{(10-10)^2+(9-10)^2+(6-10)^2+(11-10)^2+(13-10)^2+(14-10)^2+(9-10)^2+(9-10)^2+(9-10)^2+(10-10)^2}{10}}

Standard deviation s :2.144

t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}t = \frac{10-7.5}{\frac{2.144}{\sqrt{10}}}t=3.687

Df = n-1 = 10-1 = 9

T critical =t_{(df,\alpha)}=t_(9,0.05)=2.26

T calculated > T critical

So, We failed to accept null hypothesis

Hence the critical t-value for such a test given an alpha level of 0.05 is 2.26

5 0
3 years ago
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