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pav-90 [236]
3 years ago
12

Match each equation to the situation it represents. Situation Equation An adult and 5 55 children rent skates. It costs $ 4 $4do

llar sign, 4 per adult for skates and a total of $ 16 $16dollar sign, 16. It costs Mikael $ 4 $4dollar sign, 4 to make a bike bag, and he sells each for $ 5 $5dollar sign, 5. He has made a profit of $ 16 $16dollar sign, 16 so far. It was 16 ° C 16°C16, degree, start text, C, end text. It began falling 4 ° C 4°C4, degree, start text, C, end text per hour until it reached 5 ° C 5°C5, degree, start text, C, end text. 16 − 4 x = 5 16−4x=516, minus, 4, x, equals, 5 ( 5 − 4 ) x = 16 (5−4)x=16left parenthesis, 5, minus, 4, right parenthesis, x, equals, 16 4 + 5 x = 16 4+5x=164, plus, 5, x, equals, 16
Mathematics
1 answer:
maks197457 [2]3 years ago
8 0

Answer:

<u>1. C.</u>

<u>2. B.</u>

<u>3. A.</u>

Step-by-step explanation:

First situation given:

An adult and 5 children rent skates. It costs $ 4 per adult for skates and a total of $ 16.

Correct equation: C

<u>4 + 5 x = 16</u>

Second situation given:

It costs Mikael $ 4 to make a bike bag, and he sells each for $ 5. He has made a profit of $ 16, so far.

Correct equation: B

<u> ( 5 − 4 ) x = 16</u>

Third situation given:

It was 16°C. It began falling 4°C per hour until it reached 5°C.

<u>Correct equation: A</u>

<u>16 − 4 x = 5</u>

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Pada hari kantin sebanyak 800 naskah kupon telah dijual,harga senaskah kupon masing masing rm 30 dan rm 50 .jumlah wang diperole
solniwko [45]

Answer:

Step-by-step explanation:

On the day of the canteen, 800 coupons were sold, the price of each coupon was RM 30 and RM 50 respectively. The amount of money earned from the sale of coupons was RM30000. How many copies of RM30 and RM50 coupons were sold?

Let:

RM 30 = x

RM 50 = y

x + y = 800 - - - (1)

30x + 50y = 30000 - - - (2)

From (1)

x = 800 - y

Put x = 800 - y in (2)

30(800 - y) + 50y = 30000

24000 - 30y + 50y = 30000

24000 + 20y = 30000

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4 0
3 years ago
In the past, 21% of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research firm ha
Afina-wow [57]

Answer:

A sample size of 79 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

\pi = 0.21

The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.  

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.09 = 1.96\sqrt{\frac{0.21*0.79}{n}}

0.09\sqrt{n} = 1.96\sqrt{0.21*0.79}

\sqrt{n} = \frac{1.96\sqrt{0.21*0.79}}{0.09}

(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.21*0.79}}{0.09})^{2}

n = 78.68

Rounding up to the nearest whole number.

A sample size of 79 is needed.

4 0
3 years ago
What equation results from completing the square and factoring x2+2×=9
Arturiano [62]
The answer should be A. If you need work I can show you. 
5 0
3 years ago
Find maclaurin series
Mumz [18]

Recall the Maclaurin expansion for cos(x), valid for all real x :

\displaystyle \cos(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}

Then replacing x with √5 x (I'm assuming you mean √5 times x, and not √(5x)) gives

\displaystyle \cos\left(\sqrt 5\,x\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt5\,x\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^{2n}}{(2n)!}

The first 3 terms of the series are

\cos\left(\sqrt5\,x\right) \approx 1 - \dfrac{5x^2}2 + \dfrac{25x^4}{24}

and the general n-th term is as shown in the series.

In case you did mean cos(√(5x)), we would instead end up with

\displaystyle \cos\left(\sqrt{5x}\right) = \sum_{n=0}^\infty (-1)^n \frac{\left(\sqrt{5x}\right)^{2n}}{(2n)!} = \sum_{n=0}^\infty (-5)^n \frac{x^n}{(2n)!}

which amounts to replacing the x with √x in the expansion of cos(√5 x) :

\cos\left(\sqrt{5x}\right) \approx 1 - \dfrac{5x}2 + \dfrac{25x^2}{24}

7 0
2 years ago
Can someone please help me on this ASAP
OLEGan [10]
<span>(a) Triangles ABC and PQR are similar triangles since they both share 2 congruent angles and therefore their 3rd angle is also congruent. So the triangles are similar due to the AAA similarity theorem. (b) The area of triangle PQR is 1 million times larger than the area of triangle ABC. This can be shown since the area of a triangle is 1/2 base times height. You can show that the base of triangle PQR is 1000 times larger than the base of triangle ABC. And since all the sides are in proportion to each other, the height of triangle PQR is also 1000 times larger than the height of triangle ABC. And since 1000 times 1000 equals 1,000,000 or 1 million, the area of triangle PQR is 1 million times larger than triangle ABC.</span>
8 0
3 years ago
Read 2 more answers
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