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vesna_86 [32]
3 years ago
5

Graven spent $55.20 on gasoline to fill her 15 gallon tank at another gas station project spent $72.80 of gasoline to fill her 2

0 gallon tank. who paid less per gallon of gasoline explain your reasoning
Mathematics
1 answer:
den301095 [7]3 years ago
6 0

You must find the unit cost in dollars per gallon for each purchase by dividing the cost by the number of gallons.

Graven: $55.20 for 15 gallons

$55.20/(15 gal) = $3.68/gal

Project: $72.80 for 20 gallons

$72.80/(20 gal) = $3.64/gal

Answer: Project paid less per gallon.

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Show that <br><br> Sin60 / cos60 = tan 60
Lerok [7]

Answer:

  • <em><u>sin60</u></em><em><u>=</u></em><em><u>opp\hyp</u></em>
  • <em><u>cos60</u></em><em><u>=</u></em><em><u>adja\hyp</u></em>
  • <em><u>tan60</u></em><em><u>=</u></em><em><u>opp\adj</u></em>
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6 0
1 year ago
The vertex of the function’s graph is given. Find c. y = x2 + 10x + c; (-5, -27)
castortr0y [4]
Actually I didn't see the (-5, -27), the answer would be -2.
5 0
3 years ago
Help pleaseeeeeeeeeeeeeeeeeeeeeee
bixtya [17]

Answer:  \bold{(1)\ \dfrac{19,683}{64}\qquad (2)\ 16}

<u>Step-by-step explanation:</u>

(1)           (12, 18, 27, ...)

The common ratio is:

r=\dfrac{a_{n+1}}{a_n}\quad r =\dfrac{18}{12}=\boxed{\dfrac{3}{2}}\quad \rightarrow \quad r=\dfrac{27}{18}=\boxed{\dfrac{3}{2}}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=12,\  r=\dfrac{3}{2}\\\\\\Equation:\\a_n =12\bigg(\dfrac{3}{2}\bigg)^{n-1}\\\\\\\\9th\ term:\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{9-1}\\\\\\a_9=12\bigg(\dfrac{3}{2}\bigg)^{8}\\\\\\.\quad =\large\boxed{\dfrac{19643}{64}}

(2)\qquad \bigg(\dfrac{1}{16},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{2}\bigg)\\\\\\\text{The common ratio is}:\\\\r=\dfrac{a_{n+1}}{a_n}\quad  r=\dfrac{\frac{1}{8}}{\frac{1}{16}}=\boxed{2}\quad \rightarrow \quad r=\dfrac{\frac{1}{4}}{\frac{1}{8}}=\boxed{2}

The equation is:

a_n=a_o(r)^{n-1}\\\\Given:a_o=\dfrac{1}{16},\  r=2\\\\\\Equation:\\a_n =\dfrac{1}{16}(2)^{n-1}\\\\\\\\9th\ term:\\a_9=\dfrac{1}{16}(2)^{9-1}\\\\\\a_9=\dfrac{1}{16}(2)^{8}\\\\\\.\quad =\large\boxed{16}

3 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 me. I need help please.
Mice21 [21]

Answer:

The correct options are;

EFGH has 4 congruent sides

Diagonal FH bisects angles EFG and EHG

Angle FEH is congruent to angle FGH

Step-by-step explanation:

1) Given that for a reflection, we have;

The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection

Therefore;

The distance of the point E from the line HF = The distance of the point G from the line HF

Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)

We can show that from the length of a line given by the equationl = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}, that the length EH ≅ GH and EF ≅ GF

Therefore since we are given that EH = EF, we have;

EH = GH = GF = EF by the definition of congruency, which gives 4 congruent sides

2) Given that EH = GH = GF = EF and HF = FH by reflective property, we have;

ΔEHF ≅ ΔGHF

∴ ∠GHF ≅ ∠EHF by Congruent Parts of Congruent Triangles are Congruent

Similarly, ∠GFH ≅ ∠EFH

Therefore, ∠GFH = ∠EFH and ∠GHF = ∠EHF

Therefore, diagonal FH bisects angles EFG and EHG

3) Given that ΔEHF ≅ ΔGHF, we have;

Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent

7 0
3 years ago
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