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strojnjashka [21]

2 years ago

9

greta has a piece of cloth that is 9 yards long . she cuts it into pieces that are each 1/3 yards long how many pieces of cloth

dose greta have mow

1 answer:

labwork [276]2 years ago

7 0

27 peices i believe. 3 peices per yard, 9 yards. 9*3=27

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Solve the equation.<br><br> x2 − 6x − 7 = 0

zlopas [31]

Answer:

x = 7 and x = -1

Step-by-step explanation:

<u>Step 1: Factor</u>

x^2 - 6x - 7 = 0

(x - 7)(x + 1) = 0

<u>Step 2: Solve for x</u>

(x - 7)(x + 1) = 0

x - 7 = 0 and x + 1 = 0

x - 7 + 7 = 0 + 7 and x + 1 - 1 = 0 - 1

x = 7 and x = -1

Answer: x = 7 and x = -1

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3 years ago

Pls help me my grades are not good

Scilla [17]

Answer:

63

Step-by-step explanation:

8 0

3 years ago

Using the numbers 1, 2, 3, and 4 (ONLY ONCE), and the operations of adding, subtracting, multiplying, dividing, and the use of p

vazorg [7]

Step-by-step explanation:

=(4+3)×(2+1)

=7×3

=21

3 0

4 years ago

Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad battery life. Batt

BartSMP [9]

Answer:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

$\mathbf{s_ 1 =16.11}$

$\mathbf{s_2 = 7.98}$

Step-by-step explanation:

Let $x_1$ and $x_2$ be the two variables that represents the battery life in hours for talking usage and battery life in hours for internet usage respectively.

The hypothesis can be formulated as:

Null hypothesis: the variance in hours of usage for talking is not greater than the the variance in hours of usage for internet.

$H_o$ : $\sigma _1 ^2 \leq \sigma _2^2$

Alternative hypothesis: the variance in hours of usage for talking is greater than the the variance in hours of usage for internet.

$H_a : \sigma_1^2 > \sigma_2^2$

The standard deviation for the battery usage for talking is :

$\bar x_1 = \dfrac{1}{n_1} \sum x_i \\ \\ \bar x_1 = \dfrac{1}{12}(35.8 +22.4+...+35.5) \\ \\ \bar x_1 = \dfrac{241.2}{12} \\ \\ \bar x_1 =20.1$

The standard deviation Is:

$s_ 1 = \sqrt{\dfrac{1}{n_1-1}\sum (x{_1i}-\bar x_i)^2}$

$s_ 1 = \sqrt{\dfrac{1}{12-1}\sum (35.8- 20.1)^2+ (35.5-20.1)^2}$

$s_ 1 = \sqrt{259.568}$

$\mathbf{s_ 1 =16.11}$

The standard deviation for the battery life usage for the internet is :

$\bar x_2 = \dfrac{1}{n_2} \sum x_{2i}$

$\bar x_2 = \dfrac{1}{10} (24.0+12.5+36.4+...+4.7})$

$\bar x_2 = \dfrac{115}{10}$

$\bar x_2 = 11.5$

Thus; the standard deviation is:

$s_2 = \sqrt{\dfrac{1}{n_2-1}(x_{2i}- \bar x_2)^2}$

$s_2 = \sqrt{\dfrac{1}{10-1}(24-11.5)^2+(4.7-11.5)^2}$

$s_2 = \sqrt{63.60}$

$\mathbf{s_2 = 7.98}$

4 0

3 years ago

Suppose that you randomly draw one card from a standard deck of 52 cards. After writing down which card was drawn, you replace t

KengaRu [80]

Answer:

0.139

Step-by-step explanation:

We are given that

Total number of cards=52

n=18

We have to find the probability of drawing at least 7 spades

Total number of spaded in deck of 52 cards=13

Probability of getting spade= $p=\frac{13}{52}=\frac{1}{4}$

Probability, $P(E)=\frac{favorable\;cases}{total\;number\;of\;cases}$

Probability of getting no spade= $q=1-p=1-\frac{1}{4}=\frac{3}{4}$

By binomial theorem of probability

$P(X=x)=\binom{n}r}p^rq^{n-r}$

P(x $\geq 7)=1-P(x\leq 6)$

The probability of drawing at least 7 spades= $1-\sum_{r=0}^{6}\binom{18}{r}(\frac{3}{4})^{18-r}(\frac{1}{4})^r$

The probability of drawing at least 7 spades=0.139

4 0

3 years ago