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dimulka [17.4K]

3 years ago

13

The Carolina Castaways and Raleigh Rockets scored a total of 96 points. The Rockets scored one third as many points as the Casta

ways.

2 answers:

zysi [14]3 years ago

8 0

The castaways scored 32 pts

Hope it helps

11Alexandr11 [23.1K]3 years ago

4 0

I believe it's 32
96/1(1/3)
96/3
32

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For questions 1- 3 find the exact sum or difference.

7nadin3 [17]

D, Because 7/15 + 9/15 is 16/15 + 4/15 =20/15 which equals 1 5/15 which turns into 1 1/3
A, 1/21 + 2/21 = 3/21 and that equals 1/7

C, 3/14+3/12 turns into 3/14+5/14= 10/14 - 8/14= 2/14=1/7

6 0

3 years ago

Please answer this and show your work please I really need help !!!

klemol [59]

Unless you just want to give money to the ice cream truck, for 0 items, you will pay 0.

Since the price of each item is $0.77, for one item you will pay $0.77. (Is that a surprise?)

Then for two items, you will pay $0.77 for each one, or $0.77 + 0.77 = $1.54. You can also compute this by making use of the fact that multiplication is a shortcut for repeated addition. The price can be found by $0.77×2 = $1.54 = f(2).

Similarly, for 3 items, you can add $0.77 three times: $0.77 + 0.77 + 0.77 = f(3), or you can use multiplication: $0.77×3 = $2.31 = f(3).

a) Our table looks like
(x, f(x))
(0, 0)
(1, 0.77)
(2, 1.54)
(3, 2.31)

b) By now, I'm sure you have realized that to find the price of x items, you multiply 0.77 by x.
f(x) = 0.77x

8 0

3 years ago

In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence

Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - <em>α</em>)% confidence interval for population proportion <em>p</em> is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of <em>z</em> for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a <em>z</em>-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

6 0

3 years ago

-2 1/8 divided by 1 1/4

attashe74 [19]

Answer:

$-1\frac{7}{10}$

Step-by-step explanation:

Convert the mixed numbers into improper fractions: $\frac{-17}{8}$ ÷ $\frac{5}{4}$
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Simplify: $-1\frac{7}{10}$

6 0

3 years ago

solniwko [45]

The answer fam is......... 3x – 4y = 4

7 0

3 years ago