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1 year ago

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Which list correctly orders A, B, and C from least to greatest when A = –8, B = StartAbsoluteValue negative 6 EndAbsoluteValue,

and C = StartAbsoluteValue 11 EndAbsoluteValue? A, B, C C, A, B B, A, C C, B, A

2 answers:

Slav-nsk [51]1 year ago

7 0

Answer:

abc is correct

Step-by-step explanation:

vladimir2022 [97]1 year ago

3 0

Answer:

A, B, C

Step-by-step explanation:

A = -8

B = |-6| = 6

C = |11| = 11

Least to greatest, negatives are the least and between 6 and 11, 11 is the greatest.

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Figures R, S, and T are all scaled copies of one another. Figure S is a scaled copy of R using a scale factor of 3. Figure T is

MakcuM [25]

<em><u>Answer:</u></em>

1/2

1/3

6

1/6

<em><u>Step-by-step explanation:</u></em>

When we are finding the opposite of a scale factor, the scale factor is always the reciprocal. E.g. From S to R, the scale factor is 3, so from R to S, the scale factor is 1/3 so:

1. From T to S, it's the reciprocal of the scale factor (2) so 1/2

2. From S to R, it's the reciprocal of the scale factor (3) so 1/3

3. From R to T, is the from a figure to a figure to a figure so we can multiply the two scale factors (2 and 3) to get 6

4. From T to R, that's the opposite of R to T, so it's the reciprocal of the scale factor from R to T (6) so 1/6

4 0

2 years ago

Malik's solution to the equation , when , is shown below. What error did Malik make first when solving the equation ? Malik did

Marianna [84]

Answer:

Malik substituted 60 for y instead of x.

Step-by-step explanation:

According to the given situation the computation of error that Malik make first when solving the equation is shown below:-

First, we will find the value of x

$\frac{2}{5} x - 4(60) = 10$

$\frac{2}{5} x = 10 + 240$

$\frac{2}{5} x = 250$

x = 625

Now, we have

$\frac{2}{5} x - 4 y = 10$

we will solve the above equation, to find the value of y

24 - 4y = 10

4y = 24 - 10

4y = 14

$y = \frac{14}{4}$

$y = \frac{7}{2}$

So, from the above calculation, the correct option is

Malik substituted 60 for y instead of x.

3 0

1 year ago

Read 2 more answers

Grace has 1.35 pounds of strawberries, 1.4 pounds of bananas, and some apples. She has more pounds of apples than pounds of stra

satela [25.4K]

Answer:

Grace could have 1.36 pounds, 1.37 pounds, 1.38 pounds or 1.39 pounds of apples.

Step-by-step explanation:

Let a represent pounds of apples.

We have been given that Grace has 1.35 pounds of strawberries. She has more pounds of apples than pounds of strawberries.

This means that a is greater than 1.35. We can represent this information in an inequality as:

$a>1.35$

We are also told that Grace has 1.4 pounds of bananas. She has fewer pounds of apples than pounds of bananas. This means that a is less than 1.4. We can represent this information in an inequality as:

$a$

Upon combining both inequalities, we will get:

$1.35$

Therefore, Grace could have 1.36 pounds, 1.37 pounds, 1.38 pounds or 1.39 pounds of apples.

6 0

2 years ago

4(3+1/4c-1/2d) apply the distributive property to create a equivalent expression

Romashka-Z-Leto [24]

Answer iiiiiiis -2d+c+12

5 0

2 years ago

Read 2 more answers

Marketing companies have collected data implying that teenage girls use more ring tones on their cell phones than teenage boys d

ASHA 777 [7]

Answer:

The p-value here is 0.0061, which is very small and we have evidence that the girls' mean is higher than the boys' mean.

Step-by-step explanation:

We suppose that the two samples are independent and normally distributed with equal variances. Let $\mu_{1}$ be the mean number of ring tones for girls, and $\mu_{2}$ the mean number of ring tones for boys.

We want to test $H_{0}: \mu_{1}-\mu_{2} = 0$ vs $H_{1}: \mu_{1}-\mu_{2} > 0$ (upper-tail alternative).

The test statistic is

T = $\frac{\bar{X}_{1}-\bar{X}_{2}-0}{S_{p}\sqrt{1/n_{1}+1/n_{2}}}$

where

$S_{p} = \sqrt{\frac{(n_{1}-1)S_{1}^{2}+(n_{2}-1)S_{2}^{2}}{n_{1}+n_{2}-2}}$ .

For this case, $n_{1}=n_{2}=20$ , $\bar{x}_{1}=3.2$ , $s_{1}=1.5$ , $\bar{x}_{2}=2.2$ , $s_{2}=0.8$ .

$s_{p} = \sqrt{\frac{(19)(1.5)^{2}+(19)(0.8)^{2}}{38}} = 1.2021$ and the observed value is

t = $\frac{3.2-2.2}{1.2021\sqrt{1/20+1/20}} = \frac{1}{0.3801} = 2.6309$ .

We can compute the p-value as P(T > 2.6309) where T has a t distribution with 20 + 20 - 2 = 38 degrees of freedom, so, the p-value is 0.0061. Because the p-value is very small, we can reject the null hypothesis for instance, at the significance level of 0.05.

3 0

1 year ago