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3 years ago

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Initially, there were equal amount of roses and tulips at a store. Each bouquet was made with 3 roses and 4 tulips. After the bo

uquets were all made, there were 30 roses and 18 tulips left in the store. How many bouquets were made?

1 answer:

adelina 88 [10]3 years ago

8 0

Answer:There were 10 bouquets of roses made and 4.5 bouquets of tulips made

Step-by-step explanation:

You might be interested in

1 Shannon runs at a constant rate. She can run 3 miles in 26 minutes. What distance, Y in miles, can she run in 15 minutes, x?

Crazy boy [7]

Answer:

1,75 miles

Step-by-step explanation:

3 miles ______26 minutes

1 miles =>26 :3 =8,5 (8 minutes and 5 seconds)

15:8,5 =1,75 miles

7 0

2 years ago

For fun question

ZanzabumX [31]

Consider the function $f(x)=x^{1/3}$ , which has derivative $f'(x)=\dfrac13x^{-2/3}$ .

The linear approximation of $f(x)$ for some value $x$ within a neighborhood of $x=c$ is given by

$f(x)\approx f'(c)(x-c)+f(c)$

Let $c=64$ . Then $(63.97)^{1/3}$ can be estimated to be

$f(63.97)\approxf'(64)(63.97-64)+f(64)$
$\sqrt[3]{63.97}\approx4-\dfrac{0.03}{48}=3.999375$

Since $f'(x)>0$ for $x>0$ , it follows that $f(x)$ must be strictly increasing over that part of its domain, which means the linear approximation lies strictly above the function $f(x)$ . This means the estimated value is an overestimation.

Indeed, the actual value is closer to the number 3.999374902...

4 0

3 years ago

A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the

EastWind [94]

Answer:

a) $\hat p=\frac{471}{1024}=0.460$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

Part b

If we replace the values obtained we got:

$0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429$

$0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491$

The 99% confidence interval would be given by (0.429;0.491)

8 0

2 years ago

A data set consists of 2, 2, 3, 4, 5, 5, and 7. Then, 20 is added to the data set. The new value will have little effect on the

ale4655 [162]

Answer:

A data set consists of 2,2,3,4,5,5 and 7. Then, 20 is added to the data set. The new value will have the little effect on the median, but it will raise the mean significantly.

Step-by-step explanation:

Since, we all know that the measures of central tendency are:

Arithmetic Mean
Median
Mode
Geometric Mean
Harmonic Mean

Let us define Arithmetic Mean and Median.

Arithmetic mean is the sum of observations divided by the total number of observations. It is denoted by (x bar).

On the other hand, median of a distribution is that value of the variable which divides it into two equal parts.

For Arithmetic Mean:

According to the question, when data set consists 2,2,3,4,5,5 and 7, the arithmetic mean for the given data = 4

After adding 20 the given data set, the mean will be = 6

For median:

When data set consists of 2,2,3,4,5,5 and 7, the median value will be the middle value of the distribution i.e. = 4

After adding 20 to the given data set, the median will be = 4.5

Here we can say that 20 is an outlier for the given data set.

Since mean has the significant effect of the outlier because it is based on all values but the median has lesser effects of the outlier as it is based only on middle values. So whatever outlier is added to median, it will not show particular changes.

Thus , a data set consists of 2,2,3,4,5,5 and 7. Then, 20 is added to the data set. The new value will have little effect on median, but it will raise the mean significantly.

6 0

2 years ago

borishaifa [10]

The answer should be the second one y=37

6 0

2 years ago