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egoroff_w [7]
3 years ago
5

Someone HELP (sorry if the picture is poor quality)

Mathematics
1 answer:
adoni [48]3 years ago
4 0
Not sure but i would go bottom left corner answer
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Bridget went fishing with her dad. Bridget Bridget caught the first fish of the day, another was 2 more than 3 times the weight
Arturiano [62]

Answer:

Bridget's first fish =f = 5 ounces

Bridget's dad first fish weighs 17 ounces

Step-by-step explanation:

Solution:

Bridget's fishes:

f ounces

2f ounces

3f+2 ounces

1/2f ounces

3/5f ounces

Total= f +2f + (3f+2) + 1/2f +3/5f

=3f + 3f + 2 + 5f+6f/10

=6f + 2 + 11f/10

=60f+11f/10 +2

=71/10f +2

Bridget's dad fishes

3f+2

4/5f

2f+4

1/2f

Total =(3f+2) + (2f+4) + 4/5f +1/2f

=3f+2+2f+4 + 8f+5f/10

=5f + 6 + 13/10f

= 50f+13f/10 + 6

=63/10f +6

Equate the total weight

71/10f +2=63/10f +6

Collect like terms

71/10f - 63/10f =6-2

71f-63f/10 = 4

8f/10=4

Cross product

8f=40

Divide both sides by 8

f=5

Bridget's first fish =f = 5 ounces

Bridget's dad first fish = 3f +2

=3(5)+2

=15+2

=17 ounces

Bridget's dad first fish weighs 17 ounces

8 0
3 years ago
CAN SOMEONE PLEASE HELP ME WITH 13?!
Orlov [11]
I’m not sure never took that test
4 0
3 years ago
Read 2 more answers
A university researcher wants to estimate the mean number of novels that seniors read during their time in college. An exit surv
lys-0071 [83]

Answer:

Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

Step-by-step explanation:

Given - A university researcher wants to estimate the mean number

            of  novels that seniors read during their time in college. An exit

            survey was conducted with a random sample of 9 seniors. The

            sample mean was 7 novels with standard deviation 2.29 novels.

To find - Assuming that all conditions for conducting inference have

              been met, which of the following is a 94.645% confidence

              interval for the population mean number of novels read by

              all seniors?

Proof -

Given that,

Mean ,x⁻ = 7

Standard deviation, s = 2.29

Size, n = 9

Now,

Degrees of freedom = df

                                = n - 1

                                = 9 - 1

                                = 8

⇒Degrees of freedom = 8

Now,

At 94.645% confidence level

α = 1 - 94.645%

   =1 - 0.94645

  =0.05355 ≈ 0.05

⇒α = 0.5

Now,

\frac{\alpha}{2} = \frac{0.05}{2}

  = 0.025

Then,

t_{\frac{\alpha}{2}, df }  = 2.306

∴ we get

Population mean = x⁻ ± t_{\frac{\alpha}{2}, df } ×\frac{s}{\sqrt{n} }

                           = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

⇒Population mean = 7 ± 2.306 × \frac{2.29}{\sqrt{9} }

3 0
3 years ago
Read 2 more answers
PLEASE HELP GUYS/GIRLS
den301095 [7]

Answer:

3

Step-by-step explanation:

7 0
3 years ago
All integers that are exactly divisible by 2 are called what?
ivann1987 [24]

Those are called "even numbers".

Also, I guess, "multiples of 2" .


4 0
3 years ago
Read 2 more answers
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