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3 years ago

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A bag contains 10 white golf balls and 6 striped golf balls. A golfer wants to add 112 golf balls to the bag. He wants the ratio

of white to striped gold balls to remain the same. How many of each should he add?

1 answer:

expeople1 [14]3 years ago

6 0

Answer:

10:6=10*7:6*7

10:6=70:42

70+42=112

White ball 70 and striped balls 42

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PLEASE SOMEONE HELP IM SO CLOSE TO BEING DONE FOR THE YEAR <br> NO LINKS OR FILES

8_murik_8 [283]

Answer:

Step-by-step explanation:

The negative out front only tells us which way the parabola will open. That means that the first one will open upside down, since it's got a negative out front. The value of the number that serves as the leading coefficient (the number stuck to the x-squared) determines how steep or wide-open the parabola's shape is. The leading coefficient in the first one is a 1, so it opens up exactly as wide as its parent graph does. The leading coefficient in the second one, a 2, means that the parabola will be skinnier or narrower than its parent. If the leading coefficient is greater than 1, the parabola will be narrower than its parent; if the leading coefficient is greater than 0 but less than 1, the parabola will open wider than its parent.

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3 years ago

What is the factored form of

Lilit [14]

Answer:

C

Step-by-step explanation:

You can distribute the values or even plug in x=2 and see which one matches.

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3 years ago

Simplify 2X^3- 3x^2- 5x +3 / x+ 1

Maslowich

Answer:

-6x+3÷x+1

Step-by-step explanation:

2x^3-3x^2-5x+3÷x+1

2x ×2x ×2x-3x +3x-5x+3÷x+1

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4 years ago

WORTH 25 POINTS!!!!!!!!!!!!!!!!!!!!

Zepler [3.9K]

70%=70/100=7/10

what you do is shade 7 of the squares and say that is 21
now each square is worth the same
so 21/7=3
each square is worth 3

10 squares
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3 years ago

Read 2 more answers

A bank is trying to determine which model of safe to install. The bank manager believes that each model is equally resistant to

alexandr402 [8]

Answer:

So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for ?D contains 0.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution

Let put some notation

x=value for Safe 1 , y = value for Safe 2

x: 103, 90, 64, 120, 104, 92, 145, 106, 76

y: 101, 94, 58, 112, 103, 90, 140, 110, 74

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: 2, -4, 6, 8, 1, 2, 5, -4, 2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{18}{9}=2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =4.093$

The next step is calculate the degrees of freedom given by:

$df=n-1=9-1=8$

Now we need to calculate the critical value on the t distribution with 8 degrees of freedom. The value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , so we need a quantile that accumulates on each tail of the t distribution 0.025 of the area.

We can use the following excel code to find it:"=T.INV(0.025;8)" or "=T.INV(1-0.025;8)". And we got $t_{\alpha/2}=\pm 2.31$

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Now we have everything in order to replace into formula (1):

$2-2.31\frac{4.093}{\sqrt{9}}=-1.152$

$2+2.31\frac{4.093}{\sqrt{9}}=5.152$

So on this case the 95% confidence interval would be given by (-1.152;5.152).

$-1.152 < \mu_{safe1}-\mu_{safe2}$

Should we conclude that the average time it takes experts to crack the safes does not differ by model at the 5% significance level?

A) Yes, the 95% confidence interval for D contains 0.

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3 years ago