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slega [8]
3 years ago
12

Mrs. Khan is mailing 2 packages. The large package weighs 6 2 3 pounds. The small package weighs 3 1 2 pounds. Which statement B

EST describes how to find out how much heavier the large package is? A) Add 6 2 3 + 3 1 2 . B) Divide 6 2 3 ÷ 3 1 2 C) Subtract 6 2 3 – 3 1 2 . D) Multiply 6 2 3 × 3 1 2 .
Mathematics
2 answers:
vagabundo [1.1K]3 years ago
6 0

<u>Answer:</u>

Mrs. Khan is mailing 2 packages. Subtract 623 – 312 best describe how to find out how much heavier the larger package is. Hence Option C is correct

<u>Solution:</u>

Given that  

Mrs khan is mailing two packages.

Weight of larger package = 623 pounds  

Weight of small package = 312 pounds

Now we need to find how much heavier larger package compare to smaller one is.

To get this if we remove weight of smaller package from larger package , we will get how much larger package is heavier.

That is<em> Weight of larger package – Weight of smaller package</em> = 623 pounds – 312 pounds

Hence option C) subtract 623 – 312 best describe how to find out how much heavier the larger package is.

amid [387]3 years ago
3 0

Answer:

The answer is to subtract 6 2/3 - 3 1/2

Step-by-step explanation:

In order to find out how much heavier the larger package is, subtract 6

2

3

– 3

1

2

.

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(7x +6) - (5x - 4) = 180
BartSMP [9]

Answer:

x= 85

Step-by-step explanation:

(7x+6)-(5x-4)=180

There is a negative 1 between the two parenthesis

-1(5x-4)

-5x+4

Now plug it back into the equation and take out the parenthesis

7x+6-5x+4= 180

2x+10=180    Now we can solve

    -10  -10

2x= 170

2x/2 = 170/2

x= 85

I hope this helps! :)

8 0
3 years ago
Read 2 more answers
1) 2+2+2=3----- 2) -----+------ = 3×6. 3) 8+-----+=------ ×8. use addition or multiplication​
dolphi86 [110]

Step-by-step explanation:

multiplication cause gives a reasonable answer.

8 0
4 years ago
Testing radial tires from two major brands to determine if there were any differences in the expected tread life. Distance in th
Savatey [412]

Answer:

Step-by-step explanation:

Hello!

The objective is to compare the average tread life of two brands of radial tires. Two random samples of 15 tires each was taken and the distance (in thousands of miles) the tread lasted was recorded.

a. Boxplots in 2nd attachment.

As you can see both variables boxes are the same size, both of them show the equal distance between the quantiles which is compatible with a symmetrical distribution. Me sample mean (black square) is in the center of the boxes and the whiskers are also equidistant. Looking at the graphics you can assume that both study variables have a normal distribution.

b.

X₁: Distance the tire tread of a brand 1 radial tire lasts.

n₁= 15

X[bar]₁= 52

S₁= 4.61

X₂: Distance the tire tread of a brand 2 radial tire lasts.

n₂= 15

X[bar]₂= 56.93

S₂= 4.83

The objective is to test if there is any difference between the two brands, to do so you have to compare the average distance the tread lasts:

H₀: μ₁ = μ₂

H₁: μ₁ ≠ μ₂

α: 0.01

The statistic to use for this test is a student t for indepeendent samples with pooled standard deviation:

t= \frac{(X[bar]_1-X[bar]_2)(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}

Sa^2= \frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} = \frac{14*(4.61^2)+14*(4.83^2)}{15+15-2}= 22.2905

Sa= 4.72

t_{H_0}= \frac{(52-56.93)-0}{4.72*\sqrt{\frac{1}{15} +\frac{1}{15} } } = -2.86

The p-value for this test is two-tailed and you can calculate it doing the following calculations:

P(t₂₈≤-2.86) + P(t₂₈≥2.86)= P(t₂₈≤-2.86) + 1 - P(t₂₈≤2.86)= 0.0040 + (1 - 0.9960)= 0.008

p-value: 0.008

Since the p-value is less than the significance level, the decision is to reject the null hypothesis. at a 1% significance level, you can conclude that the expected of the tread life of radial tires of brand 1 is different than the expected tread life of radial tires of brand 2.

c.

You have to create a 99% CI for the difference between the population means of tread life of tires from brand 1 and 2, the formula for the CI is:

(X[bar]₁-X[bar]₂)±t_{n_1+n_2-2;1-\alpha /2}*(Sa*√(1/n₁+1/n₂))

t_{n_1+n_2-2;1-\alpha /2}= t_{28;0.995}= 2.763

(52-56.93)±2.763*(4.72*√1/15+1/15)

[-9.69;-0.17]

With a 99% confidence level, you'd expect that the interval [-9.69;-0.17] will contain the difference between the expected tread life of the radial tires of brand 1 and the expected tread life of the radial tires of brand 2.

I hope you have a SUPER day!

3 0
3 years ago
Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t
Genrish500 [490]

Answer: she deposited $2000 at 1.9%

She deposited $4000 at 3.3%

Step-by-step explanation:

Let x represent the amount that she deposited in the account earning 1.9% simple interest.

Let y represent the amount that she deposited in the account earning 3.3% simple interest.

Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account that earns 3.3% simple interest. This means that

x = y/2

The formula for simple interest is expressed as

I = PRT/100

For the savings account earning 1.9% simple interest,

I = (x × 1.9 × 1)/100 = 0.019x

For the money market account that earns 3.3% simple interest,

I = (y × 3.3 × 1) = 0.033y

If the total interest after one year is $170.00, it means that

0.019x + 0.033y = 170 - - - - - - - - - -1

Substituting x = y/2 into equation 1, it becomes

0.019×y/2 + 0.033y = 170

Cross multiplying, it becomes

0.019y + 0.066y = 340

0.085y = 340

y = 340/0.085

y = $4000

x = y/2 = 4000/2

x = 2000

3 0
3 years ago
What type of number is<br> -?<br> 2
erica [24]

Answer:

the number -2 is pronounced negative two, therefore, the type of number is a negative.

Step-by-step explanation:

8 0
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