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tamaranim1 [39]

1 year ago

13

Roger has 4 gallóns of orange juice. He puts the same amount of juice each of 5 pitchers. How many gallons of orange juice are i

n 1 pitcher?

1 answer:

salantis [7]1 year ago

5 0

Answer:

1.25

Step-by-step explanation:

you divide 5 within 4 to get how much Roger puts on each pitch

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In the diagram of △EHG below, JF ∥HG, EJ=12, JH=24, and EF=6. What is the length of EG?

Natasha_Volkova [10]

It’s 18, i guess
FG = JHxEF/EJ = 12 + 6 = 18

4 0

1 year ago

Henri bought a swimsuit at a cost of $8. Which statements are true regarding the cost of the suit? Select three options.

lora16 [44]

Answer:

The first, second to last and last stements are true

Step-by-step explanation:

You times 8 by the percentage as

e.g 8 x 1.50 = 12 is the same as marked up by 50%

e.g 8 x 1.70 =13.6 is the same as marked up by 70%

7 0

2 years ago

Read 2 more answers

An unloaded truck and trailer, with the driver aboard, weighs 30{,}00030,00030, comma, 000 pounds. When fully loaded, the truck

kari74 [83]

Answer:

1131 pounds.

Step-by-step explanation:

We have been given that an unloaded truck and trailer, with the driver aboard, weighs 30,000 pounds. When fully loaded, the truck holds 26 pallets of cargo, and each of the 18 tires of the fully loaded semi-truck bears approximately 3,300 pounds.

First of all, we will find weight of 18 tires by multiplying 18 by 3,300 as:

$\text{Weight of tires of the fully loaded semi-truck}=18\times 3,300$

$\text{Weight of tires of the fully loaded semi-truck}=59,400$

The weight of 26 pallets would be weight of 18 tires minus weight of unloaded truck.

$\text{Weight of 26 pallets of cargo}=59,400-30,000$

$\text{Weight of 26 pallets of cargo}=29,400$

Now, we will divide 29,400 by 26 to find average weight of one pallet of cargo.

$\text{Average weight of one pallet of cargo}=\frac{29,400}{26}$

$\text{Average weight of one pallet of cargo}=1130.769230769$

$\text{Average weight of one pallet of cargo}\approx 1131$

Therefore, the average weight of one pallet of cargo is approximately 1131 pounds.

3 0

2 years ago

A group of people are spinning a spinner that is separated into eight equal parts. Group One spins the spinner eight times. Grou

algol13

if the number of times spun is not a multiple of 8, it is impossible to land on an exactly even distribution for this spinner. But since the probabilities of landing on an exactly even distribution are so small, that detail doesn't perturb the overall trend, except perhaps down at n=8 spins So group 3

6 0

2 years ago

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

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

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

5 0

2 years ago