A blank is a rectangular

Mathematics

1 answer:

UNO [17]3 years ago

8 0

Answer:

I would say parallelogram. So the rectangle still fits but more shapes can be applied.

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You are considering purchasing a new home.

dexar [7]

Answer the answer is B) 990

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3 years ago

A toy company intends to ship out its entire stock of 10,254 toy trucks, and three times as many toy cars. If each box holds 35

svetoff [14.1K]

There are 10,254 toy trucks

and 10,254 * 3 = 30,762 toy cars.

In total, there are 10,254+30,762= 41,016 toys

when we divide 41,016 by 35, the quotient is the number of the full boxes, and the remainder is the number of toys left.

41,016=35*1171+31

Answer: 1171 full boxes, 31 toys left for the last box

6 0

3 years ago

A bag with 8 marbles is shown below. (2 marbles are yellow, 3 are red, and 3 are blue.) A marble is chosen from the bag at rando

Andreas93 [3]

Answer:

$\frac{5}{8}$

Step-by-step explanation:

Hi there!

We're given that there are 8 marbles in total. 2 marbles are yellow and 3 are red, making the number of red AND yellow marbles 5 (2+3=5)

Because 5 out of the 8 marbles are yellow or red, the probability of choosing one of them is $\frac{5}{8}$ .

I hope this helps!

7 0

3 years ago

Read 2 more answers

In a survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99% confidence interval for th

marysya [2.9K]

Answer:

The population proportion is estimated to be with 99% confidence within the interval (0.1238, 0.2012).

Step-by-step explanation:

The formula for estimating the population proportion by a confidence interval is given by:

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

Where:

$\hat{p}$ is the sample's proportion of success, which in this case is the people that regularly lie during surveys,

$z_{\alpha /2}$ is the critical value needed to find the tails of distribution related to the confidence level,

$n$ is the sample's size.

<u>First</u> we compute the $\hat{p}$ value:

$\hat{p}=\frac{successes}{n}=\frac{98}{603}=0.1625$

<u>Next</u> we find the z-score at any z-distribution table or app (in this case i've used StatKey):

$z_{\alpha /2}=2.576$

Now we can replace in the formula with the obtained values to compute the confidence interval:

$\hat{p}\pm z_{\alpha /2}\times\sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}=0.1625\pm 2.576\times\sqrt{\frac{0.1625\times(1-0.1625)}{603}}=(0.1238, 0.2012)$