Two times antonio's age plus three times Sarah's age equals 34. Sarah age is also five times antonio's age . How old is Sarah?

1 answer:

6 0

Let Sarah's age be s and Antonio's age be a.

According to your question,

2a + 3s = 34
s = 5a

substituting the value of s in eq. 1,

2a + 3(5a) = 34

= 2a + 15 a = 34

17a = 34

thus, a = 34/17 = 2

substituting the value of a is s,

s = 5(2)
= 10

Thus, Sarah's age is 10 years and Antonio's age is 2 years.

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Tadeo is a beekeeper. The table shows the weekly number of bees producing honey in a new hive. Which expression gives the number

Afina-wow [57]

The exponential equation that gives the number of bees in week 12 is of:

$y = 4(3)^{11} = 708588$

<h3>What is an exponential function?</h3>

It is modeled by:

$y = ab^x$

In which:

a is the initial value.

b is the rate of change.

In this problem, considering that the initial week is the first, and the value triples every week, the expression for the number of bees after x weeks is given by:

$y = 4(3)^{x - 1}$

Hence, after 12 weeks, the expression is:

$y = 4(3)^{11} = 708588$

More can be learned about exponential equations at brainly.com/question/25537936

7 0

2 years ago

Read 2 more answers

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:

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