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2 years ago

3x + 5y = 78

1 answer:

3 0

X = 6
y = 12

hope this helps at all.

you rewrite the second equation as 2x=y and then replace the y in the first equation by 2x

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Conor has a roll of fence that is 25 feet long. He needs 8 yards of fence to go around a flower bed. Does he have enough fence t

vlabodo [156]

Answer:

Yes he has enough fence to go round the flower bed.

Step-by-step explanation:

1 yard = 3 feet

25 feet = 25/3 = 8.33 yards

Yes he has enough fence to go round the flower bed.

8 0

2 years ago

Which phrase best describes the translation from the graph y = (x – 5)2 + 7 to the graph of y = (x + 1)2 – 2? A. 6 units left an

VikaD [51]

Answer:

A. 6 units left and 9 units down

Step-by-step explanation:

Given:

y = (x – 5)^2 + 7

y = (x +1)^2 - 2

The graph of y = (x – 5)^2 + 7 is a parabola 5 units right and 7 units up

The graph of y = (x +1)^2 - 2 is a parabola 1 unit left and 2 units down

Transformation is the difference in the shift

Transformation = 6 units left and 9 units down.

Therefore, answer A. 6 units left and 9 units down

Hope this will helpful.

Thank you.

4 0

2 years ago

Read 2 more answers

A pond had 60 fish. 27 were carp and the rest were tench. 10 fish were added. The probability that a fish picked at random is a

ivanzaharov [21]

Answer: 2

Explanation:

A pond had 60 fish but 10 were added so now the pond has 70 fish.

The probability of choosing a carp is 1/2 so $\frac{1}{2} (70)$ = 35 are now carp.

There were 27 carp and now there are 35 carp so 8 carp were added.

10 fish were added and 8 of them were carp so 2 of them were tench.

6 0

2 years ago

g An irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the

blagie [28]

Answer:

A 90% confidence interval of the true mean is [$119.86, $123.34].

Step-by-step explanation:

We are given that an irate student complained that the cost of textbooks was too high. He randomly surveyed 36 other students and found that the mean amount of money spent on textbooks was $121.60.

Also, the standard deviation of the population was $6.36.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean amount of money spent on textbooks = $121.60

$\sigma$ = population standard deviation = $6.36

n = sample of students = 36

$\mu$ = population mean

Here for constructing a 90% confidence interval we have used One-sample z-test statistics as we know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $121.60-1.645 \times {\frac{6.36}{\sqrt{36} } }$ , $121.60+1.645 \times {\frac{6.36}{\sqrt{36} } }$ ]

= [$119.86, $123.34]

Therefore, a 90% confidence interval of the true mean is [$119.86, $123.34].

5 0

2 years ago

a meal of a chicken sandwich and some pasta salad had 100 grams of carbohydrates the pasta salad had 10 grams more than twice th

Pie

Let the amount of carbohydrates in chicken sandwich be x

The amount of carbohydrates in salad will be 2x + 10

(x ) + ( 2x + 10) = 100

x + 2x + 10 = 100

x + 2x = 100 - 10

3x = 90

x = 90 / 3

x = 30

Therefore chicken sandwich has 30 grams of carbohydrates.

Salad will have:

2 (x) + 10 → 2 (30) + 10 → 60 + 10 = 70

Salad has 70 grams of carbohydrates.

8 0

2 years ago