Ask question

Ask question

All categories

Irina-Kira [14]

3 years ago

6

A scuba diver was at a depth of 125 feet below sea level. He descends 205 more feet below sea level. What is the new depth of th

e scuba diver?

1 answer:

EleoNora [17]3 years ago

4 0

Answer:

The new depth of the scuba diver is 330 feet below sea level.

Step-by-step explanation:

125 + 205 more below sea level = 330 below sea level

You might be interested in

Janice needs to re-shingle the roof of her house. One bundle of shingles costs $35.99 and covers 2.25 m2.

larisa [96]

To calculate the number of shingles, we first calculate for the total area of the roof of Janice's house. From the figure, we can calculate the area of the two shapes: pyramid and a rectangular prism.

A(pyramid) = lw+l√[(w/2)²+h²]+w√[(l/2)²+h²]
A = 208 m²
A(prism) = 2(wl+hl+hw)
A = 293 m²

Total area = 501 m²

Number of shingles she needs = 501 m² / 2.25 m² =<span> 223</span><span>
</span>

7 0

3 years ago

1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts o

Sauron [17]

Answer:

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A $(100-\alpha )\%$ confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a $(100-\alpha )\%$ probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the <em>t</em>-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

$\bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}$

Here,

$\bar x$ = sample mean

<em>s </em>= sample standard deviation

<em>n</em> = sample size

$t_{\alpha/2,(n-1)}$ = critical value.

The degrees of freedom for the critical value is, (<em>n</em> - 1) = 7 - 1 = 6.

The significance level is: $\alpha =1-Confidence\ level=1-0.99=0.01$

The critical value is:

$t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143$

**Use the <em>t</em>-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753$

$\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379$

The 99% confidence interval for μ is:

$x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)$

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).

6 0

3 years ago

Brainlist if you get it right

denpristay [2]

Experimental probability: 10/50

Percentage: 20%

I dont know for sure sorry if it is wrong

8 0

3 years ago

PLSSSS Plsssss help quick

Neporo4naja [7]

Answer:

(x, y) (1,5.5) , (2,5.75) (3,5.83) (4,5.875) (5,5.9)

Step-by-step explanation:

The x values lie on the horizontal line and the y values lie on the vertical line.

In order to find the value of y you must substitute the value of x in the equation given and you must plot you graph using the answers you got which I wrote for you.

NOTE:if there's anything you don't understand let me know.

5 0

3 years ago

During the 7th examination of the Offspring cohort in the Framingham Heart Study, there were 1219 participants being treated for

AlexFokin [52]

Answer:

95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

Step-by-step explanation:

We are given that during the 7th examination of the Offspring cohort in the Framing ham Heart Study, there were 1219 participants being treated for hypertension and 2,313 who were not on treatment.

The sample proportion is : $\hat p$ = x/n = 1219/3532 = 0.345

Firstly, the pivotal quantity for 95% confidence interval for the proportion of the population is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion = 0.345

n = sample of participants = 3532

p = population proportion

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level of

significance are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

<u>95% confidence interval for p</u>= [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.345-1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ , $0.345+1.96 \times {\sqrt{\frac{0.345(1-0.345)}{3532} } }$ ]

= [0.3293 , 0.3607]

Hence, 95% confidence interval for the proportion of the population which are on treatment is [0.3293 , 0.3607].

6 0

3 years ago