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3 years ago

14

Help and show steps pls

1 answer:

Ghella [55]3 years ago

4 0

You can immediately see that the vertex is located at (-3,-2) and that the y-intercept is 7 or very close to 7.

The general equation for a parabola with vertex at (h,k) is

y-k = (x-h)^2, and from that and the given info, we get:

y-(-2) = (x-(-3))^2, or y+2 = (x+3)^2, or y = (x+3)^2 - 2.

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Answer please boys????

jasenka [17]

<ABC + <ABC = 180
3x- 9 + 7x - 1 = 180
10x - 10 = 180
10x = 190
x = 19

<ABC = 3x- 9 = 3(19) - 9 = 48

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3 years ago

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What is the missing value in this table (assume a constant rate)

MrMuchimi

The answer should be B.

7 0

3 years ago

A data set includes 103 body temperatures of healthy adult humans having a mean of 98.3degreesF and a standard deviation of 0.73

faust18 [17]

Answer:

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher

Step-by-step explanation:

Data provided in the question:

sample size, n = 103

Mean temperature, μ = 98.3

°

Standard deviation, σ = 0.73

Degrees of freedom, df = n - 1 = 102

Now,

For Confidence level of 99%, and df = 102, the t-value = 2.62 [from the standard t table]

Therefore,

CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}},Mean + \frac{t\times\sigma}{\sqrt{n}})$

Thus,

Lower limit of CI = $(Mean - \frac{t\times\sigma}{\sqrt{n}})$

or

Lower limit of CI = $(98.3 - \frac{2.62\times0.73}{\sqrt{103}})$

or

Lower limit of CI = 98.11

and,

Upper limit of CI = $(Mean + \frac{t\times\sigma}{\sqrt{n}})$

or

Upper limit of CI = $(98.3 + \frac{2.62\times0.73}{\sqrt{103}})$

or

Upper limit of CI = 98.49

Hence,

CI = (98.11 , 98.49)

The value of 98.6°F suggests that this is significantly higher and the mean temperature could very possibly be 98.6°F

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3 years ago

2.9 × 109 − 1.1 × 106

padilas [110]

The answer is 33390

Hope its help ;)

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3 years ago

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Five cakes cost £3.50. How much do seven cakes cost?

Nata [24]

3.50/5 = .70
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so 7 cakes cost 4.90 pounds.

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3 years ago

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