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5 days ago

Emily and her children went into a grocery store and she bought $20.80 worth of bananas and peaches. Each banana costs $0.80 and

each peach costs $2. She bought a total of 14 bananas and peaches altogether. Write a system of equations that could be used to determine the number of bananas and the number of peaches that Emily bought. Define the variables that you use to write the system.

Mathematics

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∠UVW and ∠XYZ are complementary angles, m∠UVW=(x−10)º , and m∠XYZ=(4x−10)º .

lawyer [12517]

We recognize that two angles, ∠UVW and ∠XYZ, are complementary, which means their sum is 90°.

Their measures are given as:

∠UVW = x - 10

∠XYZ = 4x - 10

Adding these, we have:

(x - 10) + (4x - 10) = 90

Simplifying:

5x - 20 = 90

Adding 20 to both sides:

5x = 110

Dividing by 5:

x = 22

Substituting back:

∠UVW = 22 - 10 = 12°

∠XYZ = 4(22) - 10 = 78°

Therefore, the values are:

x = 22°

∠UVW = 12°

∠XYZ = 78°

6 0

3 months ago

Read 2 more answers

Supervisor: "You took 500 calls in 30 days." Representative: "That means I took an average of __________ calls per day. I'm pret

Inessa [12570]

Answer:
17 calls daily
Step-by-step explanation:
500 calls ÷ 30 days = 16.6666667 calls daily
OR
≈ 17 calls per day

4 0

1 month ago

An experiment was conducted to record the jumping distances of paper frogs made from construction paper. Based on the sample, th

tester [12383]

Answer:

$9.9676 - 2.326*0.5904 =8.594$

$9.9676 + 2.326*0.5904 =11.341$

Step-by-step explanation:

Notation

$\bar X$ is the sample mean

$\mu$ indicates the population mean (the variable of interest)

s signifies the sample standard deviation

n denotes the sample size

Solution to the problem

The mean's confidence interval is derived from the following formula:

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

In this instance, the 9% confidence interval corresponds to:

$8.8104 \leq \mu \leq 11.1248$

We can determine the mean using the following:

$\bar X = \frac{8.8104 +11.1248}{2}= 9.9676$

Additionally, the margin of error can be calculated as:

$ME= \frac{11.1248- 8.8104}{2}= 1.1572$

The margin of error for this situation is expressed as:

$ME = t_{\alpha/2}\frac{s}{\sqrt{n}} = t_{\alpha/2} SE$

Next, we find the standard error:

$SE = \frac{ME}{t_{\alpha/2}}$

The critical value for a 95% confidence interval using a normal standard distribution is roughly 1.96, and substituting gives us:

$SE = \frac{1.1572}{1.96}= 0.5904$

For the 98% confidence interval, the significance corresponds to $\alpha=1-0.98= 0.02$ and $\alpha/2 = 0.01$ with a critical value of 2.326, yielding a confidence interval of: