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3 years ago

15

A summer camp rewards campers and counselors with badges. The camp orders 200 badges. The plan to give 25 badges to counselors.

They ordered at least 3 badges for each camper. Hie many campers can there be at the camp

1 answer:

AlexFokin [52]3 years ago

8 0

Answer:

Step-by-step explanation:

200-25= 175
175÷3= 58 $\frac{1}{3}$

There are 58 campers

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4 ............I think

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2 years ago

Mya is paid $960 a week to walk 20 dogs. If she is paid the same amount to walk each dog , what is the cost to walk one dog each

Sveta_85 [38]

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2 years ago

Please can someone help?

alexandr402 [8]

Hey there! :)

Answer:

(5, -2), or x = 5 and y = -2.

Step-by-step explanation:

We can solve the two equations algebraically by eliminating a variable:

2x + 5y = 0

3x - 4y = 23

Eliminate the x variable by finding the least common multiple and multiplying both equations:

3(2x + 5y = 0)

2(3x - 4y = 23)

Distribute and subtract the bottom equation from the top:

6x + 15y = 0

6x - 8y = 46

------------------

0x + 23y = -46

23y = -46

y = -2.

Plug in y into an equation to solve for x:

2x + 5(-2) = 0

2x - 10 = 0

2x = 10

x = 5. Therefore:

The solution to this equation is (5, -2), or x = 5 and y = -2.

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3 years ago

Read 2 more answers

*PLS HELP!!*<br>{The Picture Is In The Attachment}

mash [69]

Answer:

B is 45 A is 135

Step-by-step explanation:

45 + 135 = 180

45 x 3 = 135

135 is three times as large as 45

135 and 45 are supplementary angles

6 0

2 years ago

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

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

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

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3 years ago