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Margaret [11]
3 years ago
8

If the temperature on the ground is 70 F then

Mathematics
2 answers:
harina [27]3 years ago
5 0

Answer: try to (x) each one by the answer to -0.0035 +70

Step-by-step explanation:

Inga [223]3 years ago
5 0

Answer:

  • 11000: 31.5 °F
  • 15000: 17.5 °F
  • 18000: 7 °F

Step-by-step explanation:

When you have repetitive calculations to do, it is often convenient to let a calculator or spreadsheet do them for you.

Basically, put the value of "a" into the formula and do the arithmetic.

  <u>Example</u>: for 11000 ft, -0.0035·11000 +70 = -38.5 +70 = 31.5

You might be interested in
a gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sel
adell [148]

Answer:

The price per wind chime that will maximize revenue = $ 315

Step-by-step explanation:

Given - A gift shop sells 160 wind chimes per month at $150 each. the owners estimate that for each $15 increase in price, they will sell 5 fewer wind chimes per month.

To find - Find the price per wind chime that will maximize revenue.

Proof -

Given that,

Total Wind chimes selling = 160

Price of each Wind chime = $150

Now,

Given that, for each $15 increase in price, they will sell 5 fewer wind chimes per month.

So,

Let the price = 150 + 15x

So,

Number of wind Chimes sold per month = 160 - 5x

So,

Total Revenue, R = (150 + 15x)(160 - 5x)

                             = 24000 - 750x + 2400x - 75x²

                             = 24000 + 1650x - 75x²

⇒R(x) = 24000 + 1650x - 75x²

Differentiate R with respect to x , we get

R'(x) = 1650 - 150x

Now,

For Maximize Revenue, Put R'(x) = 0

⇒1650 - 150x = 0

⇒150x = 1650

⇒x = 1650/150

⇒x = 11

∴ we get

Price per Wind chime = $ 150 + 15(11)

                                    = $ 150 + 165

                                    = $ 315

So,

The price per wind chime that will maximize revenue = $ 315

3 0
3 years ago
Three hundred ninety six divided by twenty four
Lunna [17]
Three hundred ninety six divided by twenty four = 16.5
5 0
3 years ago
Find a set of parametric equations for y= 5x + 11, given the parameter t= 2 – x
Sati [7]

Answer:

x = 2-t and y = -5\cdot t +21

Step-by-step explanation:

Given that y = 5\cdot x + 11 and t = 2-x, the parametric equations are obtained by algebraic means:

1) t = 2-x Given

2) y = 5\cdot x +11 Given

3) y = 5\cdot (x\cdot 1)+11 Associative and modulative properties

4) y = 5\cdot \left[(-1)^{-1} \cdot (-1)\right]\cdot x +11 Existence of multiplicative inverse/Commutative property

5) y = [5\cdot (-1)^{-1}]\cdot [(-1)\cdot x]+11 Associative property

6) y = -5\cdot (-x)+11  \frac{a}{-b} = -\frac{a}{b} / (-1)\cdot a = -a

7) y = -5\cdot (-x+0)+11 Modulative property

8) y = -5\cdot [-x + 2 + (-2)]+11 Existence of additive inverse

9) y = -5 \cdot [(2-x)+(-2)]+11 Associative and commutative properties

10) y = (-5)\cdot (2-x) + (-5)\cdot (-2) +11 Distributive property

11) y = (-5)\cdot (2-x) +21 (-a)\cdot (-b) = a\cdot b

12) y = (-5)\cdot t +21 By 1)

13) y = -5\cdot t +21 (-a)\cdot b = -a \cdot b/Result

14) t+x = (2-x)+x Compatibility with addition

15) t +(-t) +x = (2-x)+x +(-t) Compatibility with addition

16) [t+(-t)]+x= 2 + [x+(-x)]+(-t) Associative property

17) 0+x = (2 + 0) +(-t) Associative property

18) x = 2-t Associative and commutative properties/Definition of subtraction/Result

In consequence, the right answer is x = 2-t and y = -5\cdot t +21.

5 0
3 years ago
Ummm can someone plsss help, i need this to bring my grade up!!?
artcher [175]

Answer:

I would say it would also be 64 degrees because if you flip the HGJ to the IJG it would be the same! Hope this helps

:)

8 0
3 years ago
Read 2 more answers
A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In
lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

                        P.Q. = \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }  ~ N(0,1)

where, \hat p_1 = sample proportion of males having blood disorder= \frac{250}{1000} = 0.25

\hat p_2 = sample proportion of females having blood disorder = \frac{275}{1000} = 0.275

n_1 = sample of males = 1000

n_2 = sample of females = 1000

p_1 = population proportion of males having blood disorder

p_2 = population proportion of females having blood disorder

<em>Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.</em>

<u>So, 95% confidence interval for the difference between the population proportions, </u><u>(</u>p_1-p_2<u>)</u><u> is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < {(\hat p_1-\hat p_2)-(p_1-p_2)} < 1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

P( (\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } < (p_1-p_2) < (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} } ) = 0.95

<u>95% confidence interval for</u> (p_1-p_2) =

[(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }, (\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }]

= [ (0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }, (0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} } ]

 = [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0
3 years ago
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