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soldier1979 [14.2K]
2 years ago
12

Johnny picked 3 pounds of strawberries. That afternoon, Johnny's sister ate 2/3 of the strawberries. How many pounds of strawber

ries did Johnny's sister eat?
Mathematics
2 answers:
Katen [24]2 years ago
8 0

Answer:

She ate 2 pounds

Step-by-step explanation:

(how do you eat 2 pounds of strawberries like what)

Pie2 years ago
3 0
She ate 2 pounds! Because 2/3 is 2! Dang see that quick math!! I’m a boss
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Brenda was going to sell all of her stamp collection to buy a video game after selling half of them she changed her mind she the
BigorU [14]

Answer:

60 stamps

Step-by-step explanation:

Has 33 now. Before adding 13 stamps, she had (33-13) 30. This is half of her total, since she already sold half. To get the whole, multiply 30 by 2. 60 She started with 60

4 0
3 years ago
How many birds were at Thursday
Katarina [22]

Answer:

4 birds

Step-by-step explanation:

4 * $4.5 = $18

12 * $6.5 = $78

78 + 18 = $96

3 0
3 years ago
0.175 as a percentage
Paul [167]

Answer: 17.5%

To convert from decimal to percent, just multiply the decimal value by 100. In this example we have: 0.175 × 100 = 17.5% (answer).

➥ The ease way:

1) Move the decimal point two places to the right: 0.175 → 1.75 → 17.5.

2) Add a % sign: 17.5%

Answer: 17.5%

3 0
2 years ago
Read 2 more answers
An article gave the following summary information for fracture strengths (MPa) of n = 196 ceramic bars fired in a particular kil
Oliga [24]

Answer:

54.80 MPa to 55.92 MPa

Step-by-step explanation:

Sample mean fracture strength (x) = 55.36 MPa

Sample standard deviation (s) = 3.99 MPa

Sample size (n) = 196.

The upper and lower bounds for a 95% confidence interval are given by:

U=x+1.960*\frac{s}{\sqrt{n}} \\L=x-1.960*\frac{s}{\sqrt{n}}

The upper and lower bounds of the confidence interval are;

U=55.36+1.960*\frac{3.99}{\sqrt{196}}\\U= 55.92\\L=55.36-1.960*\frac{3.99}{\sqrt{196}}\\L= 54.80\\

The 95% confidence interval for true average fracture strength is 54.80 MPa to 55.92 MPa

4 0
4 years ago
The Paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free mailing labels an
lions [1.4K]

Answer:

Confidence interval:  (0.04649,0.04913)

Step-by-step explanation:

We are given the following in the question:

Sample size, n =  100,000

Number of people who donated, x = 4781

\hat{p} = \dfrac{x}{n} = \dfrac{4781}{100000} = 0.04781


95% Confidence interval:

\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

Putting the values, we get:

0.04781
\pm 1.96(\sqrt{\dfrac{0.04781
(1-0.04781
)}{100000}})\\\\=0.04781
\pm 0.00132\\\\=(0.04649,0.04913)

is the required 95% confidence interval  for the true proportion of their entire mailing list who may donate.

3 0
3 years ago
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