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3 years ago

Help please! asap! Thank you!

Mathematics

1 answer:

Nady [450]3 years ago

7 0

Answer:

A and E

Step-by-step explanation:

A will give the total pound.

E will give the number of servings.

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Write the number 254 as a sum of hundreds,tens,and ones

Aloiza [94]

200+50+4=254 is the answer

5 0

3 years ago

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Plz help me solve these problems

leva [86]

Answer: CAN YOU PLS MARK AS BRAINLIEST??

a.x=11/4

b.x=6

c.x=21/2

d.x=0

Step-by-step explanation:

-2+3x+x=9-x

-2+3x+x=9

4x=9+2

4x=11

x=11/4

when you move a numerator to the other side it becomes a denominator

x+2=3/4x6

x+2=3/24

x=8-2

x=6

5-2x-12=14

2x=14-5+12

2x=21

x=21/2

1/2x=-3-1/2x+4-1

1/2x+1/2x=-3+4-1

x=0

6 0

3 years ago

Which product is negative

vovangra [49]

Answer:

4th option because the two negatives will become positive and the 3rd will make it negative again

5 0

2 years ago

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When seven is subtracted from half of a number, the result is negative three. Find the number

Harlamova29_29 [7]

Answer:

Step-by-step explanation:

0.5x-7=3

add 7 to both sides

0.5x=4

multiply both sides by 2

x=8

5 0

2 years ago

Ten engineering schools in the United States were surveyed. The sample contained 250 electrical engineers, 80 being women; 175 c

steposvetlana [31]

Answer:

There is a significant difference between the two proportions.

Step-by-step explanation:

The (1 - α)% confidence interval for difference between population proportions is:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

Compute the sample proportions as follows:

$\hat p_{1}=\frac{80}{250}=0.32\\\\\hat p_{2}=\frac{40}{175}=0.23$

The critical value of z for 90% confidence interval is:

$z_{0.10/2}=z_{0.05}=1.645$

Compute a 90% confidence interval for the difference between the proportions of women in these two fields of engineering as follows:

$CI=(\hat p_{1}-\hat p_{2})\pm z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})}{n_{1}}+\frac{\hat p_{2}(1-\hat p_{2})}{n_{2}}}$

$=(0.32-0.23)\pm 1.645\times\sqrt{\frac{0.32(1-0.32)}{250}+\frac{0.23(1-0.23)}{175}}\\\\=0.09\pm 0.0714\\\\=(0.0186, 0.1614)\\\\\approx (0.02, 0.16)$

There will be no difference between the two proportions if the 90% confidence interval consists of 0.

But the 90% confidence interval does not consists of 0.

Thus, there is a significant difference between the two proportions.

4 0

2 years ago