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nordsb [41]
2 years ago
6

I ONLY NEED HELP WITH 2 AND 6 TEH REST ARE FOR INFO

Mathematics
1 answer:
Shalnov [3]2 years ago
7 0
Try one of these answers
You might be interested in
Create a system of equations
Vika [28.1K]

Answer:

Answer: Janet is 16, and David is 11.

Step-by-step explanation:

Let the ages be j and d.

j = d + 5

j + d = 27

Substitute d + 5 for j in the second equation.

d + 5 + d = 27

2d + 5 = 27

2d = 22

d = 11

Substitute 11 for d in the first equation.

j = d + 5

j = 11 + 5

j = 16

Answer: Janet is 16, and David is 11.

6 0
2 years ago
In the equation 0.75s-5/8 = 44, how do you combine the like terms?
garri49 [273]

Answer:

Step-by-step explanation:

In the given equation, the "like terms" are the constants 5/8 and 44.

It simplifies the math if we eliminate the fractions first.  Note that 0.75 = 6/8, so now we have:

8(6/8)s - 8(5/8) = 44).

Multiplying all three terms by 8 (above) yields

8(6s) - 8(5) = 8(44), or

48s              = 8(44 + 5), or 48s = 8(49)

Dividing both sides by 48 yields s:   s = 8(49/48)

Review "like terms:"  These are terms that have at least one characteristic in common.  5/8 and 44 are like terms because they are only constants (no variables are present).  We must add 5/8 and 44.   0.75s does not have a "like term" in the given equation.

6 0
3 years ago
Martin runs a 5 kilometer race in 25 minutes. What is his unit rate?
SOVA2 [1]

Answer:

[ 60/25 ] x 5 km/h = 12 km/h

Step-by-step explanation:

4 0
3 years ago
Please help!! Will give brainliest
Angelina_Jolie [31]

Answer:

b) 16 rt 2

Step-by-step explanation:

using the 45-45-90 triangle trick,

let x=16

then BC is x rt 2, or 16 rt 2

4 0
3 years ago
Suppose a simple random sample of size nequals36 is obtained from a population with mu equals 74 and sigma equals 6. ​(a) Descri
OlgaM077 [116]

Part a)

The simple random sample of size n=36 is obtained from a population with

\mu = 74

and

\sigma = 6

The sampling distribution of the sample means has a mean that is equal to mean of the population the sample has been drawn from.

Therefore the sampling distribution has a mean of

\mu = 74

The standard error of the means becomes the standard deviation of the sampling distribution.

\sigma_ { \bar X }  =  \frac{ \sigma}{ \sqrt{n} }  \\ \sigma_ { \bar X }  =  \frac{ 6}{ \sqrt{36} }  = 1

Part b) We want to find

P(\bar X \:>\:75.9)

We need to convert to z-score.

P(\bar X \:>\:75.9)  = P(z \:>\: \frac{75.9 - 74}{1} )  \\  = P(z \:>\: \frac{75.9 - 74}{1} ) \\  = P(z \:>\: 1.9) \\  = 0.0287

Part c)

We want to find

P(\bar X \: < \:71.95)

We convert to z-score and use the normal distribution table to find the corresponding area.

P(\bar X \: < \:71.95)  = P(z \: < \: \frac{71.9 5- 74}{1} )  \\  = P(z \: < \: \frac{71.9 5- 74}{1} ) \\  = P(z \: < \:  - 2.05) \\  = 0.0202

Part d)

We want to find :

P(73\:

We convert to z-scores and again use the standard normal distribution table.

P( \frac{73 - 74}{1} \:< \: z

5 0
3 years ago
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